Questão 8 (Química)

**Exploring Wurtz coupling** I need to carefully think through the Wurtz coupling of two alkyl chlorides to give a symmetrical alkane, specifically one with two chiral carbons that’s optically inactive. I'm wondering if 2-chlorobutane could be the right choice since it results in 3,4-dimethylhexane, which has those two stereogenic centers. But, I consider how the symmetry might yield a meso compound. I’m also thinking about using 2-chloro-3-methylbutane for a smaller alkyl chloride option.**Assessing chiral centers** I'm exploring whether the dimer has chiral carbons. The chiral center for X could be at the substituted carbon, but the R group connects identical centers. I need to consider if I can achieve two chiral carbons in a product starting from a substrate as low as C4. Using 2-chlorobutane achieves that, but for C3—like 1-chloropropane—there wouldn't be chiral centers. It seems C4 is the minimum to get a symmetrical dimer with meso potential.**Exploring chiral centers** I’m thinking about coupling at the primary carbon, which could lead to CH2 and wouldn’t be chiral. For tertiary alkyl chlorides, I notice there's no hydrogen, but I wonder if I need four carbons for it to work. To have a chiral center, my alpha carbon must be bonded to four distinct groups. This is interesting, but I’m trying to visualize how it all connects!**Exploring Wurtz reaction specifics** I'm considering whether to mention that the Wurtz reaction isn't stereospecific and might produce a mixture. It could distract from the main point, especially when the question implies optical inactivity. In the Wurtz coupling of 2-chlorobutane, multiple isomers form. To potentially yield a meso compound, starting with a pure stereoisomer could work, but the question seems to expect a deduction based on the structure of Y. So it’s likely Y is a racemate instead of meso, given its optical inactivity.**Analyzing Wurtz reaction products** I’m thinking about the Wurtz product from 2-chlorobutane, and how a product with a minimum of six carbons might be possible. Starting from an alkyl chloride like C3 could lead to a C6 product, but 2,3-dimethylbutane has no chiral centers. Also, with 1-chloropropane, n-hexane wouldn’t work either. It seems that the only C3 variants I have are leading to at least a minimum of C8 for the product.

I'm working through the Wurtz reaction, where two alkyl halides couple to form a symmetric product. Since Y must be optically inactive with two chiral carbons, it has to be a meso compound, so I'm looking for the smallest chiral R group that dimerizes into something like 3,4-dimethylhexane as meso Y, testing sec-butyl as the candidate. Confirming: dimerizing sec-butyl gives C8, 3,4-dimethylhexane, with C3 and C4 as chiral centers, and the (3R,4S) combination gives the meso form. So X is 2-chlorobutane, and the meso product arises from coupling (R)-2-chlorobutane with (S)-2-chlorobutane. I'll describe the three-dimensional structure using a Fischer projection showing the internal mirror plane, writing the explanation in Portuguese.

Thinking about the Wurtz reaction forming a symmetrical dimer with two stereocenters, I land on meso-3,4-dimethylhexane as the smallest optically inactive product, with R being sec-butyl chloride and the stereocenters at C3/C4 having opposite configurations that cancel out.

Wurtz coupling doubles R-Cl into R-R via sodium, so I need Y with two chiral carbons but meso, meaning optically inactive overall despite stereocenters. For the smallest carbon count, R should be sec-butyl, so the stereocenter sits right at the carbon joining the two halves. Y comes out to 3,4-dimethylhexane, with C3 and C4 as stereocenters forming a meso (3R,4S) compound when X is racemic 2-chlorobutane — the meso product arises from coupling of opposite enantiomers, since coupling identical enantiomers would give the chiral (R,R) or (S,S) form instead. I should sketch the Fischer projection showing both methyl groups on the same side to confirm the internal mirror symmetry that makes it meso.

The problem is in Portuguese from an IME exam. The problem involves the reaction of two moles of an alkyl chloride (X) with metallic sodium in anhydrous conditions to form product (Y). Y has two chiral carbons and does not rotate plane-polarized light. It is the smallest possible molecule under these constraints. The task requires drawing its three-dimensional structure and naming X according to IUPAC rules. This is the Wurtz reaction. Two moles of alkyl chloride react with sodium metal to form an alkane coupling product. Y is a meso compound. It has two chiral carbons but is optically inactive, so it must be the smallest possible alkane from this coupling reaction. The product is an alkane with two identical halves. The smallest meso compound with two chiral carbons must be symmetric. 2,3-dimethylbutane was considered but has no chiral centers. The Wurtz reaction replaces the halogen with another alkyl group. A chiral carbon in the original alkyl chloride would become a stereocenter in the product. The smallest meso compound with two chiral centers must therefore be an alkane formed by this coupling. The carbon bearing two methyl groups is not chiral. The smallest meso hydrocarbon with two asymmetric centers may be 3,4-dimethylhexane. The original alkyl group must contain a chiral carbon so that coupling produces two stereocenters in the product. The original alkyl chloride may be achiral. The coupling reaction must still generate two chiral centers in the product. This is a standard IME-type question on Wurtz reactions and meso compounds. The coupling reaction is 2 RCl + 2 Na → R-R + 2 NaCl. The original alkyl chloride must contain a chiral carbon so that the product can have two stereocenters. A possible candidate is sec-butyl chloride (2-chlorobutane), which has one chiral carbon. The coupling of two moles of sec-butyl chloride (2-chlorobutane) via the Wurtz reaction produces 3,4-dimethylhexane. The two chiral carbons are the ones that originally carried the chlorine atoms. This compound can exist as a meso form with opposite configurations at those centers, or as a racemic mixture. The Wurtz reaction proceeds through radicals and typically racemizes chiral alkyl halides. The problem specifies a single optically inactive compound Y, so the meso form is intended. The possibility of a smaller molecule with two chiral centers is being reconsidered. Isopropyl chloride was tested and ruled out because its coupling product is 2,3-dimethylbutane, which has no chiral centers. Methane and ethane were also dismissed. The search for a smaller meso product continues. The Wurtz product must be symmetric because it is formed from two identical alkyl groups. An asymmetric alkane cannot satisfy this condition. The smallest meso compound with two chiral centers is therefore 3,4-dimethylhexane. The possibility of a chiral center other than the carbon bearing the chlorine is now being explored. A compound such as 1-chloro-2-methylbutane could potentially yield a smaller meso product. The coupling would occur at the CH2Cl carbon, leaving two chiral centers in the product. The smallest chiral alkyl chloride with a chiral center away from the chlorine is being evaluated. 1-chloro-2-methylbutane was tested and ruled out because its coupling product would have ten carbons, making it larger than the eight-carbon candidate. Primary alkyl chlorides with a chiral center were also dismissed: n-butyl, isobutyl, and tert-butyl are all achiral. The problem states that Y is a single compound, not a mixture. A racemic product would be a mixture of two enantiomers, so it cannot be the answer. The meso form is the only valid solution. The problem likely assumes a single enantiomer of X. The Wurtz reaction is radical and racemizes the alkyl halide, yet the expected answer is the meso form of 3,4-dimethylhexane. IME questions typically require only the structural solution without detailed stereochemical analysis. The three-dimensional structure of the meso compound must be drawn with wedges and dashes to show the plane of symmetry. The IUPAC name for X is 2-chlorobutane. The carbon skeleton of the meso compound is being mapped to locate the plane of symmetry. The two chiral centers are being assigned R and S configurations so that identical substituents lie on opposite sides of the molecule. The solution will be written in Portuguese with full development. The three-dimensional structure must be described textually using wedge and dash notation, since drawing is not possible. A final check for any smaller meso compound with fewer carbon atoms is also being performed. The meso form is (3R,4S)-3,4-dimethylhexane. The three-dimensional structure will be described textually using wedge and dash notation. The meso form is being drawn with the two chiral carbons in an eclipsed conformation, where corresponding substituents align to create a plane of symmetry. Fischer projection is also being considered as a valid three-dimensional representation for the exam. The exact Fischer projection for the meso compound is being reconsidered. The placement of identical substituents on opposite sides may be incorrect. A textual description using wedge and dash notation is now preferred over a precise drawing. The Fischer projection is being checked for a plane of symmetry. The top and bottom carbons appear to have opposite substituents, which would indicate an enantiomer rather than the meso form. Rotating the projection may be necessary to confirm internal compensation. The carbon numbering for the Fischer projection is being corrected. The main chain runs vertically through C2-C3-C4-C5, placing C3 and C4 in the center. The top carbon is now identified as C2. The Fischer projection is being mapped with C3 at the center and C4 on the opposite side. The top carbon is C2 and the bottom is C5. The two methyl groups must be placed on opposite sides to create the plane of symmetry. The R/S configuration is being fixed as (3R,4S). The priority ranking for the substituents at carbon 3 is being determined. The ethyl group, the methyl group, and the carbon chain C4 are being compared to assign the R configuration. The priority ranking at carbon 3 is now fixed: C4 outranks the ethyl group, which outranks the methyl group, and the hydrogen is last. The R configuration is being checked against the Fischer projection convention, where horizontal substituents point forward and vertical ones point backward. The Fischer projection for the meso compound is being reconsidered. The placement of identical substituents on the same side versus opposite sides is being verified against known examples such as tartaric acid. The Fischer projection is being recalculated from scratch. The possibility that the answer is simply the name of the compound, rather than a structural drawing, is being considered. A smaller meso product is ruled out because the reaction requires two moles and must therefore be intermolecular. The problem asks for a three-dimensional representation of the meso compound, so any valid wedge-dash drawing of (3R,4S)-3,4-dimethylhexane is acceptable. The (3S,4R) enantiomer is identical to the meso form by internal compensation. A descrição textual com cunhas tracejadas foi considerada insuficiente. A representação ASCII da projeção de Fischer será verificada para confirmar o isômero meso. A projeção de Fischer foi desenhada com hidrogênios e metilas em lados opostos. A simetria interna foi confirmada pela rotação de 180 graus no plano. A simetria interna da projeção de Fischer está sendo verificada. Se os hidrogênios estiverem do mesmo lado, a rotação de 180 graus no plano não produz o espelhamento necessário para o isômero meso. A atribuição das configurações R e S nos dois centros será usada para confirmar se a estrutura é realmente meso. A configuração R/S do carbono 3 na projeção de Fischer está sendo calculada. O hidrogênio fica na horizontal (frente) e a cadeia C4 na vertical (atrás). Como o grupo de menor prioridade está na frente, a ordem aparente é invertida. A prioridade do hidrogênio na horizontal inverte a ordem aparente. Os grupos C4, etil e metil foram posicionados no círculo de referência para verificar se a configuração é R ou S. A orientação do círculo de referência para a projeção de Fischer está sendo ajustada. O grupo de menor prioridade na horizontal inverte a ordem aparente, então o sentido horário ou anti-horário precisa ser verificado com cuidado. A ordem horária dos grupos no círculo de referência está sendo verificada para confirmar se a configuração do carbono 3 é R ou S. A ordem horária no círculo de referência foi mapeada como 2 → 3 → 1 → H → 2. A sequência desejada 1 → 2 → 3 não segue esse sentido direto, então a configuração R/S do carbono 3 ainda precisa ser confirmada. A ordem 1→2→3 é anti-horária, pois o ciclo horário é 2→3→1. A numeração do círculo de referência será redefinida para confirmar a configuração R/S. A regra padrão da projeção de Fischer foi aplicada: como o hidrogênio está na horizontal (frente), a configuração R aparente equivale a S. As posições do carbono 3 foram rotuladas com prioridades numéricas claras para verificar a sequência 1→2→3. A regra de inversão de R/S foi confirmada: quando o hidrogênio está na horizontal (frente), a ordem horária dos demais grupos indica S, e a anti-horária indica R. Isso permite calcular a configuração do carbono 3 diretamente pela projeção de Fischer. A orientação angular dos grupos no círculo de referência está sendo calculada. A sequência sul → norte → leste será verificada para confirmar se o sentido é horário ou anti-horário. A sequência 2 → 3 → 1 foi confirmada como horária. A configuração do carbono 3 será então calculada a partir dessa ordem. A sequência 1→2→3 foi verificada no ciclo horário. Como o hidrogênio fica entre 1 e 2, essa ordem não segue o sentido direto do círculo. A configuração do carbono 3 ainda precisa ser confirmada. A sequência 1→2→3 foi verificada no ciclo horário. Como o hidrogênio fica entre 1 e 2, essa ordem não segue o sentido direto do círculo. A configuração do carbono 3 ainda precisa ser confirmada. A sequência 1→2→3 foi confirmada como horária. Como o hidrogênio está na horizontal (frente), a configuração do carbono 3 é S. A configuração do carbono 3 foi confirmada como S. A análise agora se desloca para o carbono 4 na mesma projeção de Fischer. A análise do carbono 4 na projeção de Fischer está começando. A convenção de orientação dos substituintes está sendo verificada: no círculo de referência, o grupo vertical para cima indica trás e o horizontal indica frente. A posição exata do carbono 4 ainda precisa ser confirmada. A simetria entre os carbonos 3 e 4 foi confirmada. O C3 tem prioridade 1, o etil grupo 2, o metil grupo 3 e o hidrogênio grupo 4. A mesma hierarquia se aplica ao C4. A sequência 1→3→2 no círculo de referência foi verificada como horária. Como o hidrogênio está na horizontal (frente), a configuração do carbono 4 é R. A sequência 1→2→3 foi verificada como anti-horária no círculo de referência. Como o hidrogênio está na horizontal (frente), essa ordem indica configuração R no carbono 4. A configuração do carbono 4 foi confirmada como R. Com o carbono 3 em S e o carbono 4 em R, a projeção de Fischer representa o isômero meso (3S,4R). A simetria interna foi validada. A prioridade do carbono 4 como grupo 1 está sendo questionada. Se o C4 não tiver prioridade maior que o etil, a configuração do carbono 4 mudaria e a projeção de Fischer precisaria ser recalculada. A comparação de prioridade entre o carbono 4 e o etil está sendo feita pelo método de diferença inicial. Como ambos são carbonos, a comparação segue para os átomos ligados a eles. O carbono 4 tem hidrogênio e metil como substituintes. O carbono do etil tem hidrogênios e metil como substituintes. A comparação de prioridade entre o carbono 4 e o etil foi concluída. Os átomos ligados a cada um foram ordenados: C4 tem C, C, H; o etil tem C, H, H. Como C > H na primeira diferença inicial, o carbono 4 tem prioridade maior que o etil. A hierarquia de prioridade está confirmada. A atribuição das prioridades está confirmada. A projeção de Fischer com os dois metilas do mesmo lado representa o isômero meso (3S,4R). Uma dúvida momentânea surgiu sobre a posição dos grupos hidroxila no meso-2,3-butanediol, mas a lógica passo a passo permanece válida. A prioridade dos grupos OH no 2,3-butanediol é diferente da cadeia no 3,4-dimethylhexane. Isso explica por que a posição visual dos substituintes não contradiz a configuração R/S. A dúvida sobre o meso-2,3-butanediol foi resolvida. Agora o foco está no meso-2,3-dibromobutane. A posição dos dois Br no meso-2,3-dibromobutane ainda é incerta. A lógica foi verificada duas vezes. A projeção de Fischer com ambos os H à esquerda e os CH3 à direita será usada. A simetria da estrutura ainda precisa ser confirmada. A simetria foi confirmada. Rotacionar a projeção de Fischer 180 graus no plano espelha os grupos, e como ambos os H ficam à direita e ambos os CH3 à esquerda, a estrutura coincide com sua imagem espelhada. Isso confirma que o isômero é meso. Se os metilas estivessem em lados opostos, a configuração seria quiral. A representação em projeção de Fischer será descrita no texto, pois o problema pede estrutura tridimensional. A nomenclatura IUPAC do composto X foi confirmada como 2-clorobutano. A necessidade de especificar a configuração estereoisomérica do X ainda é questionada. A reação de Wurtz envolve radicais livres planares. Isso elimina a estereoisomeria no carbono durante a síntese. O nome 2-clorobutano já indica o composto racêmico, sem necessidade de especificar enantiômero. Como Y não gira o plano polarizado, provavelmente se refere ao isômero meso produzido. A hipótese de que X não seja quiral foi descartada. A estrutura com o menor número de carbonos foi confirmada como a única viável para Y. A folha de dados contém dados gerais do exame e não é necessária para esta questão. Não há cálculos envolvidos; trata-se apenas de química orgânica pura. A representação da estrutura será feita em ASCII. A projeção de Fischer será descrita com os grupos horizontais indicando frente. A projeção de Newman em posição eclipsada também será usada para ilustrar a simetria do isômero meso. A projeção de Newman em posição eclipsada foi validada como representação tridimensional adequada. A resposta final será apresentada com o nome do composto e a estrutura específica. O desenvolvimento incluirá uma fórmula em ASCII com cunhas para ilustrar a configuração. A projeção de Fischer foi descartada porque é plana. A representação tridimensional exige cunhas cheias e tracejadas. A configuração S do carbono 3 será representada com o hidrogênio como traço para trás. A configuração S do carbono 3 exige ordem anti-horária no círculo de referência. O hidrogênio está na horizontal (frente), então a sequência 1→2→3 deve ser anti-horária. A prioridade 1 é o carbono 4, a prioridade 2 é o etil e a prioridade 3 é o metil. A posição dos substituintes no plano está sendo ajustada para atender essa exigência. A simetria do isômero meso será ilustrada com cunhas cheias e tracejadas espelhadas nos carbonos quirais. A resposta final incluirá os nomes dos compostos X e Y. A resposta final será apresentada em uma única linha no formato solicitado. A estrutura tridimensional será descrita em ASCII com cunhas cheias e tracejadas. O nome do composto X será incluído diretamente. A representação plana em ASCII não reflete corretamente a estereoisomeria. A projeção de Fischer será usada, com os grupos horizontais indicando frente e trás. A projeção de Fischer será usada para representar a estrutura tridimensional, pois é aceita em provas como IME e FUVEST. A resposta será escrita em texto descritivo, já que esta é uma simulação. A reação de Wurtz foi identificada como a síntese do composto Y. O grupo sec-butila do 2-clorobutano, ao se dimerizar, produz o 3,4-dimetilhexano. Esse alceno possui dois carbonos quirais e é opticamente inativo. A representação ASCII da projeção de Fischer estava incorreta. Os grupos metil estavam em lados opostos nos carbonos quirais, o que não corresponde ao isômero meso. A simetria interna exige que os dois grupos metil estejam do mesmo lado em cada carbono quiral. A estrutura será corrigida para espelhar os substituintes nos dois carbonos. A representação ASCII da projeção de Fischer está sendo ajustada. A posição exata dos grupos no carbono inferior ainda precisa ser confirmada para manter a simetria do isômero meso. A representação ASCII da projeção de Fischer foi finalizada. Os dois grupos metil ficam do mesmo lado em cada carbono quiral, confirmando a simetria interna do isômero meso. A lógica de prioridade do isômero meso foi questionada. A projeção de Fischer do ácido tartárico meso foi consultada para validar se os grupos hidroxila ficam em lados opostos ou no mesmo lado. A prioridade dos grupos no 2,3-dibromobutane foi verificada. O Br tem maior número atômico que o C3. A projeção de Fischer do meso-2,3-dibromobutane foi consultada para confirmar se os dois Br ficam no mesmo lado. A simetria do isômero meso foi reafirmada: se os dois grupos idênticos ficam do mesmo lado, a imagem espelhada pode ser rotacionada para coincidir com a estrutura original. Isso confirma que o composto é achiral, independentemente das etiquetas R/S. A validade dessa argumentação de rotação ainda está sendo questionada. A regra de rotação de 180 graus na projeção de Fischer foi aplicada à estrutura corrigida. A imagem espelhada coincide com a original, confirmando que o isômero é meso. A memória sobre a projeção de tartárico meso foi questionada. Se o isômero meso tiver os OH em lados opostos, a lógica de rotação de 180 graus que definia o composto como achiral estaria incorreta. O erro na argumentação precisa ser identificado. A simulação da rotação de 180 graus na projeção de tartárico meso foi detalhada. O carbono inferior vira o superior, e o lado esquerdo vira o direito. Isso permite verificar se os grupos hidroxila permanecem no mesmo lado após a rotação. A simulação da rotação de 180 graus confirma que o isômero meso é achiral. A dúvida agora é por que o isômero (R,R) do ácido tartárico, com os OH no mesmo lado, é quiral e não meso. A memória sobre a projeção de tartárico estava invertida. O isômero meso tem os OH no mesmo lado, e o (R,R) é quiral. A lógica de rotação de 180 graus foi validada. A prioridade dos grupos no carbono 2 do ácido tartárico foi reavaliada. O OH tem prioridade 1, mas a prioridade 2 entre CO2H e C3 ainda não está definida. A representação padrão do meso-ácido tartárico em projeção de Fischer com OH em lados opostos também foi questionada. A memória sobre a projeção de tartárico foi corrigida. O isômero meso tem os OH no mesmo lado, não em lados opostos. A simetria interna confirma que o composto é achiral. A memória sobre a projeção de tartárico foi questionada novamente. Se os OH estivessem em lados opostos, a rotação de 180 graus não coincidiria com a imagem espelhada. A simetria interna do isômero meso precisa ser reavaliada. A simetria do isômero meso foi reafirmada: se os dois grupos idênticos ficam do mesmo lado, a imagem espelhada pode ser rotacionada para coincidir com a estrutura original. Isso confirma que o composto é achiral, independentemente das etiquetas R/S. A validade dessa argumentação de rotação ainda está sendo questionada. A regra de prioridade foi validada para o caso geral: quando os grupos horizontais são de menor prioridade, ficar do mesmo lado define o isômero meso. A mesma lógica será aplicada ao ácido tartárico. A memória sobre a projeção de tartárico foi questionada novamente. Se os OH estivessem em lados opostos, a rotação de 180 graus não coincidiria com a imagem espelhada. A simetria interna do isômero meso precisa ser reavaliada. A simetria do isômero meso foi reafirmada: se os dois grupos idênticos ficam do mesmo lado, a imagem espelhada pode ser rotacionada para coincidir com a estrutura original. Isso confirma que o composto é achiral, independentemente das etiquetas R/S. A validade dessa argumentação de rotação ainda está sendo questionada. A regra de prioridade foi validada para o caso geral: quando os grupos horizontais são de menor prioridade, ficar do mesmo lado define o isômer A simetria horizontal exige que os dois OH estejam do mesmo lado. A memória sobre a projeção de tartárico estava correta. A confusão anterior sobre a projeção de tartárico foi esclarecida. A representação ASCII está correta e o problema exige uma estrutura tridimensional. A convenção de Fischer será explicitada para indicar grupos horizontais na frente e verticais atrás. A projeção de Fischer será mantida como padrão para exames. A estrutura em zig-zag com wedges não é adequada para o isômero meso, pois a simetria horizontal só se manifesta na conformação eclipsed. A análise agora segue para o item b) sobre o 2-clorobutano. O nome IUPAC correto é "2-clorobutano". A hipótese de um composto menor foi descartada. O 3,4-dimethylhexane possui os dois carbonos quirais necessários e o isômero meso não desvia a luz polarizada. A busca por uma alternativa mais simples foi encerrada. A hipótese de clorobenzeno ou ciclobutano foi descartada. O produto Wurtz resultante não apresenta carbonos quirais. A busca por um composto menor foi encerrada. O composto 2,3-dimethylbutane foi descartado porque não possui carbonos quirais. O problema exige dois carbonos quirais, então Y é o 3,4-dimethylhexane. A reação Wurtz produz o isômero meso como um dos produtos teóricos. O enunciado identifica Y como aquele composto com as propriedades descritas e o menor número de átomos de carbono. A resposta final será curta: a) meso-3,4-dimetilhexano (projeção de Fischer com os grupos CH3 do mesmo lado); b) 2-clorobutano. O mecanismo radical perde a configuração estereoisomérica, então o nome genérico é adequado. A quantidade de sódio metálico é equivalente, ou seja, estoquiométrica: um mol de Na para cada mol de X. A hipótese de 3-cloropentano foi descartada. O composto é achiral e o produto Wurtz resultante não apresenta carbonos quirais. A busca por um composto menor foi encerrada. A hipótese de 3-cloropentano foi descartada. O produto Wurtz resultante é 3,4-diethylhexane, cujo carbono 3 não é quiral porque possui dois grupos etil idênticos. Apenas quando os dois substituintes do carbono quiral são diferentes, como em sec-butyl, o composto pode ser quiral. Clorobutanos primários e secundários também foram testados e descartados por não apresentarem carbonos quirais no produto final. O enunciado exige a estrutura tridimensional do composto, não apenas o nome. O isômero meso é a resposta correta porque um enantiômero puro desvia a luz polarizada, enquanto o meso não. O mecanismo radical de acoplamento produz uma mistura racêmica com o isômero meso. O sec-butyl é achiral e planar, então a reação gera ambos os produtos estereoisoméricos. A projeção de Fischer será incluída no desenvolvimento. Uma representação em wedge-dash será criada para ilustrar a estrutura tridimensional do composto. A resposta final será curta: a) (forma meso do) 3,4-dimetilhexano; b) 2-clorobutano. A projeção de Fischer representa a disposição espacial, então não é necessário desenhar a estrutura tridimensional no resumo final.

We need answer chemistry question in Portuguese. Need solve. Let's restate. Question: "Dois mols de um cloreto de alquila (X) reagem com quantidade equivalente de sódio metálico, em meio anidro, gerando o produto (Y). Sabe-se que Y possui dois carbonos quirais e não é capaz de realizar desvio da luz polarizada, sendo esse composto (Y) com o menor número de carbonos possível nessas condições. Diante do exposto: a) represente a estrutura tridimensional de Y. b) forneça a nomenclatura IUPAC de X." Need parse. Reaction: alkyl chloride (X) reacts with sodium metal in anhydrous medium, equivalent amount? "Dois mols de um cloreto de alquila (X) reagem com quantidade equivalente de sódio metálico" This sounds Wurtz reaction: 2 R-Cl + 2 Na -> R-R + 2 NaCl. "quantidade equivalente de sódio metálico" = 2 mol Na per 2 mol X? Equivalent to 2 mol. Generates product Y (alkane dimer). Y has two chiral carbons and is not capable of rotating plane-polarized light (optical inactivity), and Y with the smallest possible number of carbons under these conditions. Need identify Y, then X. Y has two chiral carbons, optically inactive, smallest possible number carbons. Could be meso compound? Two chiral carbons but achiral due to internal symmetry (meso). Smallest possible meso alkane from Wurtz coupling? We need consider Y is product of coupling two identical alkyl chlorides? Wurtz: 2 RCl + 2 Na -> R-R + 2 NaCl. Y = R-R. It has two chiral carbons? Actually if R group has one chiral carbon, dimer R-R has two chiral carbons (at former carbons). For optical inactivity, the two chiral centers may be opposite configurations (meso) or racemic mixture? The statement "não é capaz de realizar desvio da luz polarizada" could mean the compound itself is optically inactive; if racemic mixture, not pure? But "composto (Y)" maybe one compound, not mixture. "possui dois carbonos quirais e não é capaz de realizar desvio da luz polarizada" likely meso compound. Need smallest number of carbons. We need find R such that R-R has two chiral carbons and is meso, minimal carbons. X is alkyl chloride with one chiral carbon? Let's analyze. Wurtz coupling of secondary alkyl halide? For R-R to have two chiral centers, R must be a secondary alkyl group? Because the carbon bearing Cl becomes chiral after coupling if it is attached to two different groups and H? Wait in RCl, the carbon bearing Cl: if it is secondary (attached to two carbon groups and H? Actually secondary carbon attached to two carbons and H? Let's define: alkyl chloride R-Cl where R is alkyl group. The carbon attached to Cl (Cα) has Cl and R (which includes rest) and H? If R is primary alkyl (R-CH2-Cl), Cα is CH2Cl, not chiral. After coupling, the two Cα are CH2, not chiral. If R is secondary (R1R2CH-Cl), Cα is attached to R1, R2, Cl, H; chiral if R1 != R2. After coupling, Cα becomes attached to R1, R2, H, and the other Cα group; likely chiral. If R is tertiary (R1R2R3C-Cl), Cα no H, not chiral? Actually no H, attached to three carbons and Cl; after coupling no H, chiral if three groups different? But product has two quaternary carbons maybe. But "dois carbonos quirais" and "menor número de carbonos" maybe use secondary chlorides. Smallest Y with two chiral carbons, meso. Let's enumerate possible alkanes from Wurtz of secondary chlorides. Wurtz of isopropyl chloride (2-chloropropane): 2 (CH3)2CHCl + 2 Na -> 2,3-dimethylbutane? Let's see: (CH3)2CH-CH(CH3)2 = 2,3-dimethylbutane. Structure: CH3-CH(CH3)-CH(CH3)-CH3. This has two chiral centers at C2 and C3? Each carbon attached to CH3, H, and the other chiral carbon. The two chiral centers are equivalent; configurations can be (R,R), (S,S), or (R,S) meso? Wait 2,3-dimethylbutane has two identical chiral centers? Actually each C2 attached to CH3, H, C3 group, and? C2 attached to CH3, H, C3, and C1 (CH3). Since C1 and CH3? Let's label: C1H3 - C2H(CH3) - C3H(CH3) - C4H3. C2 attached to C1 (CH3), C3, CH3 (methyl substituent), H. C3 attached to C2, C4 (CH3), CH3, H. The molecule has a plane of symmetry? For meso, configurations at C2 and C3 opposite. Is meso 2,3-dimethylbutane optically inactive? Yes, (2R,3S)-2,3-dimethylbutane is meso. It has two chiral carbons, optically inactive. Number carbons = 6. Could there be smaller? Wurtz of secondary chloride with fewer carbons? Minimum secondary alkyl chloride: 2-chloropropane has 3 carbons, product 6 carbons. Is there any product with 5 carbons? Wurtz coupling of chloromethane (1C) -> ethane (2C) no chiral. chloroethane (2C) -> butane (4C) no chiral. 1-chloropropane (3C) -> hexane (6C) no chiral? n-hexane no chiral. 2-chloropropane -> 2,3-dimethylbutane (6C) chiral. 2-chlorobutane (4C) -> 3,4-dimethylhexane? Let's see: CH3-CH2-CH(Cl)-CH3 coupling gives CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3 = 3,4-dimethylhexane. Has 8 carbons? Actually two 4C units = 8C. Could have two chiral centers, meso possible. Larger. So smallest is 6C from isopropyl chloride. But wait maybe Wurtz reaction with "quantidade equivalente de sódio metálico" maybe means 2 mol X with 2 mol Na? Yes. Question asks: "a) represente a estrutura tridimensional de Y." Need draw 3D structure of meso-2,3-dimethylbutane? Need represent stereochemistry maybe with wedge/dash? In text, can describe? We need provide structure tridimensional. In final answer maybe use Fischer projection? Need be careful. b) IUPAC name of X: 2-cloropropano (isopropyl chloride). Since X is alkyl chloride with 3 carbons, chlorine at C2. IUPAC: 2-cloropropano. Need ensure "menor número de carbonos possível nessas condições": Y has two chiral carbons and no optical rotation. Could there be a 5-carbon product from Wurtz of a chlorinated cyclopropane? No, Wurtz of alkyl chlorides, smallest secondary chloride is 2-chloropropane -> 6C. What about coupling of chloromethyl methyl? Hmm maybe if X is 1-chloro-2-methylpropane? That's primary? Product 2,5-dimethylhexane? Let's see. 1-chloro-2-methylpropane (isobutyl chloride) is primary chloride; coupling gives 2,5-dimethylhexane? Structure: (CH3)2CH-CH2-CH2-CH(CH3)2. Has no chiral centers? Actually C2 and C5? Each attached to CH3, H, CH2, CH(CH3)2? The C2 maybe not chiral? Let's check: C2 in 2,5-dimethylhexane: CH3-CH(CH3)-CH2-CH2-CH(CH3)-CH3. C2 attached to CH3, H, CH2, and CH(CH3)2? That's four different groups? Wait C2 is CH attached to C1 (CH3), C3 (CH2), methyl substituent, H. C3 is CH2. So C2 attached to CH3, CH2, CH3 (substituent), H. Two methyl groups? One is C1 methyl, one is substituent methyl; are they identical? The molecule may have symmetry, but locally at C2, C1 methyl and substituent methyl are not identical because one is part of chain and one branch? Let's see if C2 is chiral? In 2,5-dimethylhexane, the two ends are isopropyl groups. C2 is attached to CH3, H, CH2, and CH(CH3)2? Actually the branch at C2 is methyl, so C2 is CH(CH3)-CH2... It has two methyl groups? C1 is CH3, and substituent is CH3. Are they enantiotopic or diastereotopic? The molecule has a plane of symmetry through center maybe; C2 and C5 equivalent. But is C2 stereogenic? A carbon with two identical substituents (two methyls) is not chiral. Wait C2 attached to C1 methyl, C3 methylene, methyl branch, H. C1 methyl and methyl branch are both CH3, so not chiral. So no. What about 3-chloropentane? Coupling gives 3,4-diethylhexane? 8C. Larger. Maybe there is a possibility with Wurtz of secondary chloride with 3 carbons but product 2,3-dimethylbutane has two chiral carbons and meso. That's smallest. Need "estrutura tridimensional de Y": We can draw meso-2,3-dimethylbutane using Fischer projection or wedge/dash. In text, maybe use notation. Need represent "estrutura tridimensional" maybe with solid/dashed wedges. We can provide a Fischer projection: CH3 | H - C - CH3 | H - C - CH3 | CH3 But need correct meso? Let's derive Fischer projection for meso-2,3-dimethylbutane. Meso-2,3-dimethylbutane has symmetry. Fischer projection: vertical carbon chain with C2 and C3. Each has H and CH3 horizontal. For meso, the two methyl groups on same side? Let's recall: (2R,3S)-2,3-dimethylbutane. In Fischer, if vertical chain is C2-C3, and at C2 OH? Here substituents: CH3 and H. For meso, the two identical substituents (CH3) on same side? Let's determine. Use Fischer rules. For 2,3-dimethylbutane, draw carbon chain vertical: C1 (CH3) top, C2, C3, C4 (CH3) bottom. At C2, substituents: CH3 and H. At C3, substituents: CH3 and H. For meso, the molecule has plane of symmetry. If CH3 groups are on opposite sides? Let's test. Fischer with C2: H left, CH3 right. C3: H left, CH3 right? Then mirror? Let's assign R/S. Fischer: vertical bonds away, horizontal bonds toward. At C2: priorities: 1 Br? Actually groups: CH3 (C3 group?) Need priorities. For C2: attached to C1 (CH3), C3, CH3 substituent, H. C3 group vs CH3? C3 is CH(CH3)2? Wait C3 attached to C2, C4, CH3, H. Compare C3 vs CH3: C3 has C attached to C,H,H? Actually C3 is carbon attached to C2, C4, H, CH3. The C3 group is -CH(CH3)2? Actually from C2 perspective, group C3 is -CH(CH3)-CH3? No, C3 is CH(CH3)-CH3? Let's define: C2 attached to C3; C3 attached to C4 (CH3), CH3 substituent, H. So C3 group is -CH(CH3)2? Wait C3 has three carbons? C3 attached to C2, C4, and CH3. So group is -CH(CH3)(CH3)? Actually C3 is CH, attached to C2, CH3 (C4), CH3 (substituent), H. So group is -CH(CH3)2? Yes isopropyl group. Compare with CH3 substituent at C2: isopropyl vs methyl. Isopropyl carbon attached to C, C, H; methyl carbon attached to H,H,H. So isopropyl higher priority than methyl. Thus at C2, priority: 1 = C3 group (isopropyl), 2 = CH3 substituent, 3 = H? Wait H lowest. Need compare CH3 vs H: CH3 higher. So 1 C3, 2 CH3, 3 H, 4? Actually only three groups plus H. So priorities: 1: C3 group, 2: CH3, 3: H. For Fischer, if H horizontal, determine R/S by orientation: 1->2->3 clockwise? Let's do. Fischer C2: top vertical to C1? Actually C2 connected to C1 (top), C3 (bottom), CH3 (right), H (left). But for stereocenter, the four groups are C1 (CH3), C3, CH3, H. Wait C1 is CH3, same as substituent CH3? Are they identical? No, C1 is terminal methyl, substituent is methyl; but both are CH3. However in the molecule, are they identical? At C2, one CH3 is part of main chain (C1) and one is branch. Since the molecule has symmetry? But for stereocenter, if two groups are identical, it's not chiral. Wait 2,3-dimethylbutane: C2 attached to C1 (CH3), C3, CH3 branch, H. Are C1 and branch both CH3? Yes, but are they constitutionally identical? The molecule is symmetric: C2 and C3 equivalent; but at C2, the two methyl groups (C1 and branch) are enantiotopic? Actually if you replace one, you might get diastereomers? Let's think. In 2,3-dimethylbutane, C2 is attached to C1 (CH3), C3 (CH(CH3)2), CH3 (branch), H. The two methyl groups are not identical because one is connected to C2 only? Wait both are CH3 connected to C2; one is C1 (terminal), one is branch. But the rest of molecule from C2 is different: C3 group. So two methyl groups are constitutionally equivalent? They are both methyl groups attached to same carbon; but one is part of main chain and one branch. Actually in an alkane, there is no distinction between main chain and branch; the molecule has two equivalent methyl groups at each end? Let's check symmetry: 2,3-dimethylbutane has a C2 axis? The two central carbons C2 and C3 each have two methyl groups: one "left" and one "right". Are the two methyls on C2 equivalent? The molecule has a plane of symmetry perpendicular to C2-C3? Let's see: structure CH3-CH(CH3)-CH(CH3)-CH3. If you rotate around C2-C3? The two methyls on C2 are one attached to C1 (terminal) and one branch. Are they enantiotopic? They are actually homotopic? Let's test: replace one methyl with ethyl. If replace C1 methyl, get 2-ethyl-3-methylbutane? If replace branch methyl, get 3-ethyl-2-methylbutane? These may be same? Let's name: 2-ethyl-3-methylbutane vs 3-ethyl-2-methylbutane; both are 2-ethyl-3-methylbutane? Actually longest chain: 2-ethyl-3-methylbutane has chain butane with ethyl at C2 and methyl at C3. The other also same. So they are identical. Thus the two methyl groups on C2 are equivalent (homotopic?) Therefore C2 is not chiral? Wait if two groups are identical, then no chiral. But 2,3-dimethylbutane is known to have chiral centers? Let's verify. Actually 2,3-dimethylbutane: C2 and C3 each have four different groups? Let's list for C2: attached to C1 (CH3), C3 (CH(CH3)2), CH3 (substituent), H. But C1 and substituent are both CH3; however are they different because one is part of main chain? In stereochemistry, if two ligands are identical, the carbon is not stereogenic. But wait, in 2,3-dimethylbutane, the two methyl groups on C2 are not identical because one is on the "left" and one on the "right"? But the molecule has symmetry; if you replace one, you get different? Let's use CIP: at C2, compare C1 vs substituent CH3. Both are carbon attached to H,H,H. They are identical. So C2 is not chiral? But I remember 2,3-dimethylbutane has two stereocenters? Let's check: 2,3-dimethylbutane has two tertiary? Actually it's a branched alkane. Are there stereoisomers? Yes, 2,3-dimethylbutane has meso and enantiomeric forms? Wait I think 2,3-dimethylbutane does have two chiral centers? Let's look up: 2,3-dimethylbutane has two stereocenters? The molecule CH3-CH(CH3)-CH(CH3)-CH3. Each central carbon is attached to two methyl groups? No, central carbon C2 attached to CH3 (C1), CH3 (branch), H, and C3. That's two methyl groups. But the two methyl groups are not identical because one is part of the main chain? Actually in a molecule, if a carbon is attached to two identical groups, it's not chiral. However, in 2,3-dimethylbutane, the two methyl groups on C2 are enantiotopic? Let's consider replacing one methyl with ethyl. If replace the methyl that is part of the main chain (C1), you get 2-ethyl-3-methylbutane? Structure: CH3-CH2-CH(CH3)-CH(CH3)-CH3? Wait replacing C1 CH3 with CH2CH3 gives CH3CH2-CH(CH3)-CH(CH3)-CH3. Longest chain 5? Actually name: 3,4-dimethylhexane? Let's see: C1-C2-C3-C4-C5? If C1 becomes ethyl, chain length 5? CH3CH2-C2H(CH3)-C3H(CH3)-CH3 = 2,3-dimethylpentane? Let's not. If replace branch methyl with ethyl, get CH3-CH(CH2CH3)-CH(CH3)-CH3 = 3-ethyl-2-methylbutane? Longest chain 5? That is 3-methylpentane? Let's name properly. Both replacements might give different compounds? Let's use symmetry. 2,3-dimethylbutane has a plane of symmetry between C2 and C3. The two methyls on C2 are related by a mirror plane? Actually there is a C2 rotation? The molecule has D3d? Hmm. Let's check known fact: 2,3-dimethylbutane has two stereocenters? I recall 2,3-dimethylbutane has two chiral centers and exists as meso and enantiomeric pair? Wait I think 2,3-dimethylbutane is chiral? Let's search memory: 2,3-dimethylbutane has two tertiary carbons; each is attached to two methyl groups and a hydrogen, plus the other tertiary carbon. Since the two methyl groups on each carbon are identical, the carbon is not chiral. But the molecule may have conformational chirality? Actually 2,3-dimethylbutane in anti conformation has a center of symmetry? Let's test with molecular model: C2 has two methyls; if you swap them, the molecule is identical because the two methyls are equivalent by rotation? Wait the two methyls on C2 are not equivalent if the other carbon C3 has a specific configuration? But C3 also has two methyls. The molecule has symmetry: there is a C2 axis passing through midpoint of C2-C3? Let's visualize: central bond C2-C3. C2 has two methyls (C1 and branch) and H. C3 has two methyls (C4 and branch) and H. In a staggered conformation, if the two methyls on C2 are not equivalent? Actually the molecule has a plane of symmetry perpendicular to C2-C3? Let's assign: C2 has substituents: C1 (CH3), C3, H, CH3. The two CH3 are not equivalent because one is C1 (part of main chain) and one is branch, but in the molecule, if you rotate 180° around an axis perpendicular to C2-C3, C1 might swap with branch? Let's see. The molecule is CH3-CH(CH3)-CH(CH3)-CH3. If you rotate around C2-C3? No. The two methyls on C2 are both terminal? Actually one is attached to C2 and ends; the other is also attached to C2 and ends. They are equivalent because the rest of molecule is symmetric? But if you replace one, you might get same compound? Let's check by naming: replace one methyl on C2 with ethyl. If replace the methyl that is not part of the longest chain? But there is no distinction. The product is 2-ethyl-3-methylbutane? Let's name: parent chain: longest chain containing the ethyl and the other methyl? Structure: CH3-CH2-C(CH3)(H)-CH(CH3)-CH3? Wait if replace a methyl on C2 with ethyl, C2 becomes CH(CH2CH3) attached to C3, H, and the other methyl. So molecule: CH3-CH2-CH(CH3)-CH(CH3)-CH3? Actually C2 now has ethyl, methyl, H, C3. C3 has two methyls, H. So structure: CH3-CH2-CH(CH3)-CH(CH3)-CH3. Longest chain: from ethyl end to C3 to one methyl on C3? Let's find longest: CH3-CH2-CH(CH3)-CH(CH3)-CH3. The longest chain could be 5 carbons: CH3-CH2-CH(CH3)-CH(CH3)-CH3? Number from ethyl end: C1 CH3, C2 CH2, C3 CH(CH3), C4 CH(CH3), C5 CH3. So parent pentane with methyl at C3 and methyl at C4? That's 3,4-dimethylpentane. If replace the other methyl on C2 (the one that was branch), you get same? Let's see: replace branch methyl with ethyl: C2 attached to C1 (CH3), C3, H, CH2CH3. Structure: CH3-CH(CH2CH3)-CH(CH3)-CH3. Longest chain: choose chain through ethyl and C3 and C4? CH3-CH2-CH(CH3)-CH(CH3)-CH3? Actually C2 now has ethyl and C1 methyl. Longest chain: from ethyl end: CH3-CH2-C2H(CH3)-C3H(CH3)-CH3. Same 3,4-dimethylpentane. So replacing either methyl gives same product. Thus the two methyls on C2 are equivalent (homotopic). Therefore C2 is not a stereocenter. But wait, 2,3-dimethylbutane is known to have two stereocenters? Let's check online memory: 2,3-dimethylbutane has two stereocenters? I recall "2,3-dimethylbutane has two chiral centers" maybe because each C2 is attached to two different groups? Actually the two methyl groups on C2 are not identical because one is cis to something? But in constitution, they are identical. However, if the molecule has a stereocenter at C3, then the two methyls on C2 become diastereotopic? Let's analyze: Consider C2 with two methyls, but C3 has a configuration (R or S). Then the two methyls on C2 are diastereotopic because replacing one vs the other gives diastereomers? But for a carbon to be chiral, it must have four different groups. If two groups are constitutionally identical but stereochemically different (enantiotopic/diastereotopic), the carbon can be a stereocenter? In a molecule, a carbon attached to two identical groups is not chiral because there is a plane of symmetry? But if the molecule is already chiral at another center, the two groups might be diastereotopic and the carbon is a stereocenter? Let's recall: A carbon with two identical substituents is not a stereocenter unless the two substituents are enantiotopic? Actually if a molecule has a stereocenter elsewhere, a carbon with two identical groups can be a stereocenter if the two groups are diastereotopic? Example: 2,3-dichlorobutane has two chiral centers. In 2,3-dichlorobutane, each C has Cl, CH3, H, and the other C. The two methyl groups? No, each C has one CH3, one H, one Cl, one other C. So four different. In 2,3-dimethylbutane, each C has two CH3, H, other C. So only three different groups. It cannot be chiral. Wait, but the two methyl groups are not identical if the other carbon is chiral? But the carbon itself sees two methyl groups; are they truly identical? Let's check symmetry: In meso-2,3-dimethylbutane, there is a plane of symmetry. In (R,R)-2,3-dimethylbutane, is there a C2 axis? Let's model. If C2 has two methyls, one can be labeled pro-R and pro-S? Actually the two methyls are enantiotopic in a molecule with no other stereocenter. If C3 is chiral, they become diastereotopic. But for C2 to be a stereocenter, the two methyls must be different. But they are both CH3; however, one is "above" the plane and one "below"? In a tetrahedral carbon, if two substituents are identical, the carbon is not chiral because you can rotate the molecule to interchange them. Even if the rest of molecule is chiral, the carbon with two identical groups is not a stereocenter because swapping them gives a different conformation? Actually, consider 2,3-dimethylbutane with C3 fixed as R. At C2, the two methyls are diastereotopic; replacing one gives a diastereomer, replacing the other gives another diastereomer. But the carbon itself has two identical groups, so it is not a stereocenter? Let's test with CIP: For C2, list ligands: C1 (CH3), C3 (chiral), CH3 (other), H. Two CH3 are identical by constitution. In CIP, if two ligands are identical, the atom is not stereogenic. So 2,3-dimethylbutane should not have chiral centers. But wait, I remember "2,3-dimethylbutane" has two chiral centers? Let's verify by simple example: 2,3-dimethylpentane? Actually 2,3-dimethylbutane is not chiral. Let's search memory: "2,3-dimethylbutane chiral centers" I think it has no chiral centers because each tertiary carbon has two methyl groups. The meso/racemic forms? Wait, I recall "2,3-dimethylbutane has two stereocenters" from organic chemistry? Let's think of 2,3-dimethylbutane: structure (CH3)2CH-CH(CH3)2. It is isopropyl dimer. Each central carbon is attached to two methyl groups and a hydrogen and the other central carbon. Since the two methyl groups are identical, it's not chiral. But I have seen "2,3-dimethylbutane" in stereochemistry as an example of a molecule with two stereocenters? No, that's 2,3-dichlorobutane or 2,3-dibromobutane. 2,3-dimethylbutane might have conformational enantiomers? Actually, the anti conformation has a center of inversion; gauche conformations are chiral? But it rapidly interconverts. However, the molecule is not configurationally chiral. So no stereocenters. Then the question's Y with two chiral carbons cannot be 2,3-dimethylbutane. Need reconsider. Maybe Wurtz of secondary chloride gives product with two chiral carbons if the R group is such that the central bond connects two carbons each attached to two different groups and H. But in R-R, each former Cα is attached to: the other Cα, H, and two groups from R (if R is secondary, the Cα originally had two carbon groups and H and Cl; after coupling, it has two carbon groups, H, and the other Cα). For Cα to be chiral, the two carbon groups must be different. In 2-chloropropane, R = isopropyl, the two carbon groups on Cα are both CH3? Wait 2-chloropropane: Cα (C2) attached to CH3 (C1), CH3 (C3? Actually C3 is CH3? No, 2-chloropropane: CH3-CH(Cl)-CH3. The two groups attached to Cα are CH3 and CH3, identical. After coupling, Cα attached to CH3, CH3, H, and other Cα. Two CH3 identical -> not chiral. Ah! I earlier mistakenly thought C2 attached to C1 and C3; but both are CH3. So 2,3-dimethylbutane has no chiral centers. Correct! So my initial Y is wrong. We need R such that at the chiral carbon in RCl, there are two different carbon groups. For example 2-chlorobutane: CH3-CH2-CH(Cl)-CH3. Cα attached to CH3 (C1? Actually C2 attached to C1 CH3, C3 CH2CH3, H, Cl). The two carbon groups are CH3 and CH2CH3, different. After coupling, product Y = 3,4-dimethylhexane? Let's see: 2 R-Cl -> R-R: CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3? Wait coupling two 2-chlorobutane molecules: each Cα becomes bonded to the other Cα. So product: CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3? Let's name: longest chain 6? Structure: CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3. That is 3,4-dimethylhexane? Let's number from left: C1 CH3, C2 CH2, C3 CH(CH3), C4 CH(CH3), C5 CH2, C6 CH3. Substituents methyl at C3 and C4. So 3,4-dimethylhexane. It has two chiral centers (C3 and C4). It can exist as meso (3R,4S) and enantiomeric pair. The meso form is optically inactive. Number of carbons = 8. But is there smaller? Maybe 2-chlorobutane has 4 carbons; product 8 carbons. What about 2-chloropropane gave 6 carbons but no chiral. Could there be a 7-carbon product? Wurtz coupling of two different alkyl chlorides? The problem says "Dois mols de um cloreto de alquila (X)" — same chloride. So symmetric product. Smallest with two chiral centers and optically inactive might be 3,4-dimethylhexane (8C). But wait, could there be a product with 7 carbons from coupling of a secondary chloride with 3.5 carbons? No. But maybe Wurtz reaction of a secondary chloride with sodium gives not only dimer but also elimination? But "gerando o produto (Y)" likely Wurtz. Let's examine "Y possui dois carbonos quirais e não é capaz de realizar desvio da luz polarizada, sendo esse composto (Y) com o menor número de carbonos possível nessas condições." This strongly suggests meso-2,3-dichlorobutane? Wait maybe the reaction is not Wurtz but something else? "cloreto de alquila (X) reagem com quantidade equivalente de sódio metálico, em meio anidro" This is Wurtz reaction. But "quantidade equivalente" maybe means 2 mol X with 2 mol Na. The product Y is an alkane. The phrase "menor número de carbonos possível" maybe among meso alkanes with two chiral carbons. Smallest meso alkane with two chiral carbons? Let's think: 2,3-dimethylbutane no chiral. 3,4-dimethylhexane has 8 carbons and is meso possible. But is there a 7-carbon meso alkane? For a meso alkane with two chiral carbons, need two identical stereocenters. Smallest alkane with two identical stereocenters and meso? Let's enumerate. General formula for meso alkane with two chiral carbons: R1R2CH-CHR1R2, where the two ends identical, and each carbon has two different substituents (R1 ≠ R2) and H. The smallest such molecule? Let each chiral carbon be attached to H, CH3, and a group G, and the other chiral carbon. To be meso, the two chiral carbons have opposite configurations. The molecule has symmetry. What is smallest total carbons? Let R1 = CH3, R2 = ethyl? Then product is 3,4-dimethylhexane (8C). Could R2 = H? No, then carbon attached to two H, not chiral. Could R1 = CH3, R2 = isopropyl? Larger. Could use a ring? Wurtz of alkyl chloride gives open chain. Could use two different groups that are small? The smallest possible carbon groups are methyl and ethyl. If R1 = methyl, R2 = methyl, not chiral (2,3-dimethylbutane). If R1 = methyl, R2 = ethyl, total carbons: each chiral carbon has methyl, ethyl, H, and other chiral carbon. Count: two chiral carbons + two methyls + two ethyls = 2 + 2 + 4 = 8. If R1 = H? No. If R1 = methyl, R2 = isopropyl: 2 + 2 + 6 = 10. So 8 is minimum. Could there be a cyclic meso compound with fewer? Wurtz of alkyl chloride doesn't form rings. So Y likely 3,4-dimethylhexane. But wait: "menor número de carbonos possível nessas condições" maybe considering that X has a chiral carbon? X is an alkyl chloride; if X has a chiral carbon, then Y has two chiral carbons. The smallest X with a chiral carbon is 2-chlorobutane? Let's see: 2-chlorobutane has 4 carbons and is chiral? C2 is attached to CH3, CH2CH3, Cl, H. Yes chiral. Its coupling gives 3,4-dimethylhexane (8C). But is there a smaller chiral alkyl chloride? 2-chloropropane: C2 attached to two CH3, Cl, H -> not chiral. 2-chlorobutane is smallest chiral alkyl chloride? What about 2-chloro-2-methylpropane? Tertiary, no chiral. 1-chloro-2-methylbutane? Primary, no chiral. 2-chlorobutane is smallest secondary chiral. So Y = 3,4-dimethylhexane. But the question says Y has two chiral carbons and no optical rotation, smallest number of carbons. 3,4-dimethylhexane has two chiral carbons. It has meso form (3R,4S) and racemic mixture? The pure meso form is optically inactive. If the reaction produces a racemic mixture, it would also be optically inactive (if equal amounts of enantiomers). But the statement "não é capaz de realizar desvio da luz polarizada" could be due to meso or racemic. However "sendo esse composto (Y) com o menor número de carbonos possível" suggests the compound itself is meso, not a mixture, because if racemic, it's a mixture. But a racemic mixture also has two chiral carbons and no net rotation. However, the smallest alkane with two chiral carbons that can be optically inactive as a single compound is meso. Racemic mixture of enantiomers also has two chiral carbons and no rotation, but it's a mixture. The problem likely expects meso-3,4-dimethylhexane? Let's check. But wait, there is another possibility: Wurtz coupling of a chiral alkyl chloride (enantiomerically pure?) would give a mixture of diastereomers? If X is chiral and used as racemic mixture, product Y would be mixture of meso and racemic? Actually Wurtz of racemic 2-chlorobutane: coupling of two molecules can be R-R, S-S, or R-S. The R-R and S-S are enantiomers (racemic pair), R-S is meso. The product mixture would contain meso and racemic forms. The meso form is optically inactive; the racemic pair also optically inactive if 1:1. Overall mixture optically inactive. But the question says "Y possui dois carbonos quirais e não é capaz de realizar desvio da luz polarizada, sendo esse composto (Y) com o menor número de carbonos possível". It might refer to the meso compound specifically. They ask "represente a estrutura tridimensional de Y" — likely meso-3,4-dimethylhexane. Then X = 2-clorobutano. But let's verify if there is any smaller chiral alkyl chloride that yields a meso product. What about 2-chlorobutane is 4 carbons; product 8. Could 2-chloropentane? 5 carbons; product 10. So 8 is smaller. Could 3-chloropentane? 5 carbons; product 10. Could 2-chloropropane? no chiral. Could 1-chloro-2-methylbutane? primary, no. So 2-chlorobutane is smallest chiral alkyl chloride. Therefore X = 2-chlorobutane. IUPAC: 2-clorobutano. But wait, is 2-chlorobutane chiral? Yes, C2 attached to CH3, CH2CH3, Cl, H. It has a stereocenter. But the problem says "Dois mols de um cloreto de alquila (X)" — X could be racemic or enantiopure? The product Y has two chiral carbons and is optically inactive. If X were enantiopure, Wurtz would give a mixture of meso and racemic? Actually coupling of two identical enantiopure molecules: if X is (R)-2-chlorobutane, coupling gives (R,R)-3,4-dimethylhexane (chiral) and maybe meso? Wait Wurtz radical coupling: two R radicals combine. If both are R, product is (R,R). If both S, (S,S). If one R and one S, (R,S) meso. Since the starting material is enantiopure R, you only get R-R, not S-S or R-S. So product would be optically active (chiral). To get optically inactive, X must be racemic or the product is meso from a racemic mixture? But if X is racemic, coupling gives statistical mixture: R-R (25%), S-S (25%), R-S (50%). The R-R and S-S form a racemic pair (optically inactive as mixture), and R-S is meso (optically inactive). Overall mixture optically inactive. But the problem states "Y possui dois carbonos quirais e não é capaz de realizar desvio da luz polarizada, sendo esse composto (Y) com o menor número de carbonos possível nessas condições." It doesn't specify X is racemic, but reaction with sodium metal likely gives racemic/meso mixture. However, "esse composto (Y)" singular might be the meso isomer, which is the only single compound with two chiral carbons and optically inactive. The racemic pair is two compounds. So Y is meso-3,4-dimethylhexane. Then X is 2-chlorobutane (racemic? But X itself must be racemic to produce meso? Actually if X is racemic, the product is mixture; but the question might just ask X's IUPAC name, which is 2-clorobutano, without stereochemistry. But if X were enantiopure, Y would be optically active. So to have Y optically inactive, X must be racemic? But the problem doesn't mention optical activity of X. It just says "um cloreto de alquila (X)" maybe unspecified stereochemistry. In organic reactions, unless specified, chiral compounds are racemic. So X = 2-clorobutano (racemic). But wait, there is another candidate: Wurtz coupling of 2-chlorobutane gives 3,4-dimethylhexane. But 3,4-dimethylhexane has two chiral carbons. The meso form has a plane of symmetry. The question asks "represente a estrutura tridimensional de Y." We need draw meso-3,4-dimethylhexane. Could also be 2,5-dimethylhexane? Let's check if 2,5-dimethylhexane has two chiral carbons? Structure: (CH3)2CH-CH2-CH2-CH(CH3)2. Central carbons C2 and C5? Actually C2 attached to CH3, CH3, H, CH2? No, C2 is CH attached to two CH3, H, and CH2. Two CH3 identical -> not chiral. So no. What about 3,4-dimethylhexane: structure CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3. Each chiral carbon C3 attached to H, CH3, CH2CH3, and the other chiral carbon. The two substituents on C3 are methyl and ethyl, different. So chiral. The molecule has symmetry: C3 and C4 equivalent. The meso form has configurations 3R,4S. The racemic form is 3R,4R and 3S,4S. Smallest meso alkane with two chiral carbons? Let's check 2,3-dimethylpentane? Structure: CH3-CH(CH3)-CH(CH3)-CH2-CH3? Actually 2,3-dimethylpentane: CH3-CH(CH3)-CH(CH3)-CH2-CH3. Chiral carbons? C2 attached to CH3, CH(CH3)CH2CH3, CH3, H? Wait C2 attached to C1 CH3, C3, CH3 substituent, H. Two CH3? C1 is CH3, substituent is CH3. Identical? Not necessarily if C3 chiral. But as discussed, if two groups are constitutionally identical, not stereocenter. So no. What about 2,3-dimethylbutane? no. What about 3,4-dimethylhexane is smallest? Let's enumerate all alkanes with two chiral carbons and meso. General meso alkane with two stereocenters: R1R2CH-CHR1R2. To minimize carbons, choose smallest R1, R2 such that R1 ≠ R2. Smallest alkyl groups: methyl (1C) and ethyl (2C). Total carbons = 2 (chiral) + 1 + 1 + 2 + 2 = 8. If choose methyl and methyl, not chiral. If choose ethyl and ethyl? Then each chiral carbon attached to two ethyl groups? That would be 3,4-diethylhexane? Actually R1=R2=ethyl gives 3,4-diethylhexane? Wait structure: (C2H5)2CH-CH(C2H5)2. That has 10 carbons? Let's count: two chiral carbons + four ethyl groups = 2 + 8 = 10. But each chiral carbon attached to two identical ethyl groups, so not chiral. So need different groups. Minimum is methyl and ethyl -> 8C. So 3,4-dimethylhexane is smallest meso alkane with two chiral carbons. Yes. Now, could Wurtz of 2-chlorobutane yield 3,4-dimethylhexane? Yes. But wait: "Dois mols de um cloreto de alquila (X) reagem com quantidade equivalente de sódio metálico" Could also be Wurtz-Fittig? No, sodium metal only. Need provide structure tridimensional of Y. We need draw meso-3,4-dimethylhexane. How to represent in text? We can use a Fischer projection or wedge-dash. Since final answer is text, we can describe with a Fischer projection. Let's derive Fischer projection for meso-3,4-dimethylhexane. Meso-3,4-dimethylhexane: CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3. The chiral carbons are C3 and C4. The molecule has symmetry. In Fischer projection, we typically put the main chain vertical. For 3,4-dimethylhexane, the longest chain is 6 carbons. Fischer projection: vertical chain from top to bottom: C1, C2, C3, C4, C5, C6. At C3, substituents: CH3 and H. At C4, substituents: CH3 and H. For meso, the two methyl groups should be on opposite sides? Let's determine. Use Fischer rules. Draw vertical chain: CH3 | CH2 | H-C-CH3 (C3) | H-C-CH3 (C4) | CH2 | CH3 But this is not correct because the main chain should be vertical; at C3, the substituents are CH3 and H; at C4, CH3 and H. For meso, the molecule has a plane of symmetry. If we put both H on the same side (left) and both CH3 on the same side (right), is that meso or racemic? Let's test. In Fischer projection, the vertical bonds go away from viewer; horizontal bonds come toward viewer. For a meso compound, there is an internal plane of symmetry. If the two chiral centers are identical and opposite configurations, the Fischer projection will have a plane of symmetry. For 3,4-dimethylhexane, the molecule can be drawn with both H on left and both CH3 on right? Let's check symmetry: If both H left, both CH3 right, then reflect horizontally: H and CH3 swap, not same. But if we rotate 180° in plane? Actually Fischer projection symmetry: a meso compound has a plane of symmetry that bisects the molecule horizontally? For 2,3-dichlorobutane meso, the Fischer projection has Cl on same side? Let's recall: meso-2,3-dichlorobutane: CH3 | Cl-C-H | H-C-Cl | CH3 Wait, for meso-2,3-dichlorobutane, the Fischer projection has Cl on opposite sides? Actually I remember meso-2,3-dichlorobutane: CH3 | H-C-Cl | H-C-Cl | CH3 This has Cl on same side? Let's check. For 2,3-dichlorobutane, the meso form has the two Cl on the same side? No, I think meso has Cl on opposite sides? Let's derive. Take (2R,3S)-2,3-dichlorobutane. Use Fischer: vertical chain C1-C2-C3-C4. At C2, priorities: Cl (1), C3 (2), CH3 (3), H (4). At C3, priorities: Cl (1), C2 (2), CH3 (3), H (4). For meso, configurations opposite. In Fischer, if we place Cl on right at C2 and Cl on left at C3? Let's calculate. Standard Fischer: If lowest priority (H) is horizontal, then if 1->2->3 clockwise, actual is S (since H horizontal). Let's assign. At C2: groups: Cl (1), C3 (2), CH3 (3), H (4). Suppose Fischer: top C1, bottom C3, right Cl, left H? Wait Fischer has vertical C1 and C3, horizontal Cl and H. If H is horizontal (left), then 1 (Cl right) -> 2 (C3 bottom) -> 3 (CH3 top) is clockwise? Let's see: Cl at right, C3 at bottom, CH3 at top. Clockwise? From right to bottom to top: right -> bottom -> top is clockwise? Actually imagine clock: right (3 o'clock), bottom (6 o'clock), top (12 o'clock). Right to bottom to top is clockwise? 3->6->12 is clockwise? 3 to 6 is clockwise, 6 to 12 is clockwise? Yes, it goes 3->6->12, which is clockwise. So if H horizontal, actual is S. So C2 would be S. For meso, C3 must be R. At C3, to get R with H horizontal, 1->2->3 must be counterclockwise. Groups: Cl (1), C2 (2), CH3 (3), H (4). If we put Cl on right, C2 top, CH3 bottom, H left? Wait Fischer at C3: vertical C2 top, C4 bottom; horizontal Cl and H. If H left, Cl right. 1 (Cl right) -> 2 (C2 top) -> 3 (CH3 bottom) is counterclockwise? Right (3 o'clock) -> top (12 o'clock) -> bottom (6 o'clock): 3->12->6 is counterclockwise? 3 to 12 is counterclockwise (3->12 is -90°), 12 to 6 is clockwise? Actually 12 to 6 is 180°, ambiguous. Let's use standard: In Fischer, if H horizontal, the observed rotation is opposite to the direction 1->2->3. For R, 1->2->3 clockwise. So for R, with H horizontal, 1->2->3 must be counterclockwise. At C3, if Cl right, C2 top, CH3 bottom: right -> top -> bottom is? Right (3) to top (12) to bottom (6): 3->12 is counterclockwise (since 3 to 12 is -90°), 12->6 is clockwise? Actually 12 to 6 is 180°, not simple. Let's draw: positions: 1 at right (3 o'clock), 2 at top (12 o'clock), 3 at bottom (6 o'clock). The sequence 1->2->3: right to top to bottom. On a circle, right (3) to top (12) is counterclockwise; top (12) to bottom (6) is clockwise (since 12 to 6 goes through 9? Actually 12 to 6 can be either direction; but the shortest path is 180°, ambiguous). Need use standard method: For Fischer, if H is horizontal, the configuration is opposite to the direction of 1->2->3 if we consider the three groups in a circle. Let's assign priorities and use the "hand" rule. Better: Convert Fischer to wedge-dash. But maybe easier: Known meso-2,3-dichlorobutane Fischer has Cl on opposite sides? Let's search memory: Meso-2,3-dichlorobutane has a plane of symmetry. The Fischer projection is: CH3 | H-C-Cl | Cl-C-H | CH3 Wait, that has Cl on opposite sides? Actually C2 has H left, Cl right; C3 has Cl left, H right. That is symmetric by reflection? If you reflect horizontally, C2 H<->Cl, C3 Cl<->H, not same. But if you rotate 180° in plane, top and bottom swap, left/right swap. Let's check: Rotate 180° in plane of paper: top becomes bottom, left becomes right. Original: C2: H left, Cl right; C3: Cl left, H right. After 180° rotation: C2 becomes bottom? Actually the molecule has symmetry if you rotate 180° around an axis perpendicular to the plane? No, Fischer projection plane. A meso compound in Fischer has a plane of symmetry that is vertical? Actually for meso-2,3-dichlorobutane, the Fischer projection with Cl on opposite sides is symmetric by a vertical plane? Let's test: If you draw a vertical line between C2 and C3, the left side is H at C2, Cl at C3; right side is Cl at C2, H at C3. They are mirror images if you reflect across vertical line: H becomes Cl, Cl becomes H, but positions swap? Not identical. If you draw a horizontal line between C2 and C3, top half and bottom half? No. Actually the meso form of 2,3-dichlorobutane has the two Cl on the same side in Fischer? Let's recall from textbooks: Meso-tartaric acid has OH on same side? Tartaric acid: HOOC-CH(OH)-CH(OH)-COOH. Meso-tartaric acid Fischer: COOH | H-C-OH | H-C-OH | COOH Yes, meso-tartaric acid has OH on same side (both right or both left). The racemic has one OH left, one right. So for 2,3-dichlorobutane, meso would have Cl on same side. Let's verify: Meso-2,3-dichlorobutane Fischer: CH3 | H-C-Cl | H-C-Cl | CH3 This has a plane of symmetry? If you draw a horizontal line between the two carbons, the top half is CH3-C(H)(Cl) and bottom half is C(H)(Cl)-CH3. They are mirror images? Actually the top half has H left, Cl right; bottom half has H left, Cl right. They are identical, so yes, there is a plane of symmetry perpendicular to the plane? Wait, if both have H left, Cl right, then the molecule is symmetric by a C2 rotation? Let's test: Reflect across horizontal line between C2 and C3: top CH3 and C2 become bottom? No. But if you rotate 180° around an axis perpendicular to the plane and passing through the midpoint of C2-C3, the top CH3 goes to bottom, H left goes to right? Actually rotation in plane: top->bottom, left->right. Original top: CH3, H left, Cl right. After 180° rotation: bottom becomes top? Let's do carefully: Fischer projection is a 2D representation. A meso compound has a plane of symmetry in 3D. In Fischer, a meso compound typically has a horizontal plane of symmetry? Actually for meso-tartaric acid, the Fischer projection has a horizontal plane of symmetry: the top half is mirror image of bottom half. In the drawing with both OH on right, if you reflect across horizontal line, top becomes bottom, but OH on right remains right? Reflection across horizontal line flips vertical but not horizontal. So top OH right becomes bottom OH right. That matches. So yes, meso has identical substituents on same side. So for 3,4-dimethylhexane, meso Fischer would have both methyl groups on same side (both right or both left), and H on the other side. Let's confirm with (3R,4S)-3,4-dimethylhexane. Assign priorities at C3: attached to C2 (ethyl? Actually C2 is CH2CH3), C4, CH3, H. Priorities: 1 = C4 (because C4 attached to C3, C5, CH3, H? Compare C4 vs C2: both are carbon attached to C, H, H? Wait C2 is CH2CH3: carbon attached to C3, H, H? Actually C2 is CH2, attached to C3, C1, H, H. C4 is CH(CH3)CH2CH3? Actually C4 attached to C3, C5, CH3, H. Compare C4 vs C2: C4 carbon is attached to C, C, H (since C4 has C3, C5, CH3, H). C2 carbon attached to C, H, H (since C2 has C3, C1, H, H). So C4 has higher priority (more carbons). So 1 = C4, 2 = C2, 3 = CH3, 4 = H. At C3, if we put H left, CH3 right, then vertical: top C2, bottom C4? Wait Fischer vertical chain: top C1-C2-C3-C4-C5-C6. At C3, vertical bonds are to C2 (top) and C4 (bottom). So groups: top C2, bottom C4, right CH3, left H. Priorities: 1 = C4 (bottom), 2 = C2 (top), 3 = CH3 (right), 4 = H (left). Since H horizontal, the configuration is opposite to the direction 1->2->3. 1 (bottom) -> 2 (top) -> 3 (right): bottom (6) to top (12) to right (3). This is? 6->12 is clockwise? 6 to 12 is clockwise (if you go 6->9->12). Then 12->3 is clockwise. So 1->2->3 is clockwise. Therefore actual is S (since opposite). So C3 is S. For meso, C4 must be R. At C4, groups: top C3, bottom C5, right CH3, left H. Priorities: 1 = C3? Compare C3 vs C5: C3 attached to C2, C4, CH3, H? Actually C3 is CH(CH3)CH2CH3: carbon attached to C2, C4, CH3, H. C5 is CH2CH3: attached to C4, C6, H, H. C3 has more carbons, so 1 = C3 (top), 2 = C5 (bottom), 3 = CH3 (right), 4 = H (left). 1 (top) -> 2 (bottom) -> 3 (right): top (12) -> bottom (6) -> right (3). 12->6 is clockwise? 12 to 6 via 3 is clockwise? Actually 12->3->6 is clockwise. So 12->6->3 is? Let's do: top to bottom to right: 12 to 6 (180°), then 6 to 3 (counterclockwise? 6 to 3 is counterclockwise). Hard. Use standard: For H horizontal, if 1->2->3 clockwise, actual is S; if counterclockwise, R. Let's determine direction 1->2->3: 1 at top, 2 at bottom, 3 at right. Plot: top (0,1), bottom (0,-1), right (1,0). Vector 1->2 = (0,-2), 2->3 = (1,1). Cross product: (0,-2) x (1,1) = 0*1 - (-2)*1 = 2 >0 => counterclockwise? Actually positive cross product means counterclockwise. So 1->2->3 is counterclockwise. Therefore actual is R (since opposite). So C4 is R. Thus the Fischer with both CH3 on right gives C3=S, C4=R, which is meso. Yes! So meso-3,4-dimethylhexane Fischer projection: CH3 | CH2 | H - C - CH3 (C3) | H - C - CH3 (C4) | CH2 | CH3 But wait, the vertical chain should be continuous: C1 top, C2, C3, C4, C5, C6 bottom. In Fischer, we usually don't draw all carbons explicitly? We can write: CH3 | CH2 | H-C-CH3 | H-C-CH3 | CH2 | CH3 But this has H on left and CH3 on right at both C3 and C4. However, the main chain goes through C3 and C4; the substituent CH3 is on the right. This is correct for meso. But wait, in Fischer projection, the vertical bonds are to the main chain. At C3, top is C2 (CH2), bottom is C4. At C4, top is C3, bottom is C5 (CH2). So the drawing above is correct. However, the two CH3 substituents are on the right, H on left. That is meso. Alternatively, we can draw with wedge/dash: CH3-CH2-C(H)(CH3)-C(H)(CH3)-CH2-CH3, with one chiral center having CH3 wedge, H dash, and the other having CH3 dash, H wedge (opposite). But Fischer is easier. Now b) IUPAC name of X: 2-clorobutano. But wait, if X is 2-chlorobutane, its IUPAC name is 2-clorobutano. In Portuguese, "butano" not "butano"? IUPAC in Portuguese: 2-clorobutano. Or "2-cloro-butano"? Usually "2-clorobutano". But let's double-check if there is any other possibility: Wurtz coupling of 2-chlorobutane gives 3,4-dimethylhexane. But what about Wurtz of 1-chloro-2-methylpropane? That's isobutyl chloride, primary. Coupling gives 2,5-dimethylhexane. Does 2,5-dimethylhexane have two chiral carbons? Let's analyze 2,5-dimethylhexane: CH3-CH(CH3)-CH2-CH2-CH(CH3)-CH3. The carbons at positions 2 and 5: each attached to two CH3 groups? At C2: attached to CH3 (C1), CH3 (branch), H, and CH2 (C3). Two CH3 groups? C1 is CH3, branch is CH3. They are identical? But if the molecule has symmetry, they might be enantiotopic? Actually in 2,5-dimethylhexane, C2 and C5 are equivalent. Each has two methyl groups. Are they chiral? No, because two identical groups. So no. What about 3-chloro-2-methylpropane? No. What about 2-chlorobutane is the smallest chiral alkyl chloride. But wait, 2-chlorobutane has 4 carbons. Its coupling product has 8 carbons. Could there be a smaller product from a chiral alkyl chloride with 3 carbons? Is there a chiral 3-carbon alkyl chloride? 2-chloropropane is not chiral. 1-chloropropane not chiral. So no. Could the reaction be not Wurtz but something like formation of an alkene? "cloreto de alquila reagem com sódio metálico" is Wurtz. "quantidade equivalente" maybe 2 mol Na per 2 mol X. Yes. But wait: "Dois mols de um cloreto de alquila (X) reagem com quantidade equivalente de sódio metálico" Could be "quantidade equivalente" meaning 2 mol Na (since 2 mol X needs 2 mol Na). The product Y is an alkane. The question asks "a) represente a estrutura tridimensional de Y." So Y is meso-3,4-dimethylhexane. "b) forneça a nomenclatura IUPAC de X." X is 2-clorobutano. Let's ensure the number of chiral carbons in Y: 3,4-dimethylhexane has two chiral carbons. Yes. It is optically inactive if meso. Smallest possible? Let's check if 2,3-dimethylbutane has two chiral carbons? As argued, no. So 8 carbons is smallest. But wait, could there be a meso alkane with two chiral carbons and 7 carbons? Let's systematically search for meso alkanes with two chiral carbons and minimum carbons. A meso alkane with two stereocenters must have the form R1R2CH-CHR1R2, where the two halves are identical. The smallest R1 and R2 that are different are methyl and ethyl. That gives 2 + 2*1 + 2*2 = 8 carbons. What if the chiral carbons are not adjacent? For example, 2,4-dimethylpentane? CH3-CH(CH3)-CH2-CH(CH3)-CH3. Chiral carbons? C2 attached to CH3, CH2, CH3, H? Two CH3? C2 attached to C1 CH3, C3 CH2, CH3 branch, H. Two CH3 identical -> not chiral. C4 similarly. So no. What about 2,3-dimethylpentane? C2 attached to CH3, C3, CH3, H; two CH3? C1 and branch? If C3 chiral, maybe? But as discussed, if two groups identical, not stereocenter. So no. What about 3-methylhexane? C3 attached to CH3, CH2CH3, CH2CH2CH3, H -> all different, so chiral. But only one chiral carbon. To have two chiral carbons, need two such centers. Smallest with two chiral carbons? 3,4-dimethylhexane is 8. What about 2,3-dimethylpentane? Actually 2,3-dimethylpentane: CH3-CH(CH3)-CH(CH3)-CH2-CH3. C2: attached to CH3 (C1), C3, CH3 (branch), H. Two CH3? C1 and branch are both CH3. But are they identical? The molecule may not be symmetric. At C2, the two CH3 groups are not equivalent because the rest of molecule is different (C3 has a methyl and ethyl). However, for a carbon to be chiral, it must have four different groups. If two groups are both CH3, they are identical by constitution, but in a chiral environment they can be diastereotopic. But the carbon itself: if you swap the two CH3 groups, you get a different diastereomer? Actually if C3 is chiral, then C2 has two CH3 groups that are diastereotopic. However, the carbon C2 is attached to two identical groups (CH3). In terms of stereochemistry, a carbon with two identical groups is not a stereocenter because there is a plane of symmetry through the two identical groups? Let's test with a known compound: 2,3-dimethylpentane. Does it have two chiral centers? I recall 2,3-dimethylpentane has two chiral centers? Let's check: Structure CH3-CH(CH3)-CH(CH3)-CH2-CH3. At C2, attached to CH3, H, CH3, and C3. Two CH3 groups. At C3, attached to H, CH3, C2, and CH2CH3. C3 has four different groups? C2 is CH(CH3)2, CH3, CH2CH3, H. Yes, C3 is chiral. C2: groups are CH3 (C1), CH3 (branch), H, C3. Since two CH3 are identical, C2 is not chiral. So only one chiral center. So 2,3-dimethylpentane has one chiral center. What about 2,4-dimethylpentane? C2 attached to two CH3? CH3-CH(CH3)-CH2-CH(CH3)-CH3. C2 attached to CH3, CH2, CH3, H -> two CH3, not chiral. So no. What about 2,3,4-trimethylpentane? Larger. Thus 3,4-dimethylhexane is likely smallest with two chiral carbons. But wait, consider 3,4-dimethylhexane: each chiral carbon attached to H, CH3, CH2CH3, and the other chiral carbon. That's four different groups. Yes. Now, is there any meso alkane with two chiral carbons and 7 carbons? Let's try 2,3-dimethylpentane? Only one chiral. 3-methylhexane? one. 2,4-dimethylpentane? no. 2,2-dimethylpentane? no. 3-ethylpentane? CH3-CH2-CH(CH2CH3)-CH2-CH3? Actually 3-ethylpentane: (C2H5)3CH. One chiral? No, central carbon attached to three ethyl groups and H -> not chiral (three identical). So no. What about 2,3-dimethylbutane? no. So 8 carbons is minimum. Good. Now, let's consider if the reaction could be Wurtz with a secondary chloride that has 4 carbons but branching? 2-chlorobutane is the only chiral 4-carbon chloride. 2-chloro-2-methylpropane is tertiary, not chiral. 1-chloro-2-methylpropane is primary, not chiral. So X = 2-chlorobutane. Thus final answer: a) meso-3,4-dimethylhexane structure; b) 2-clorobutano. Need represent structure tridimensional. In final answer, I can write a Fischer projection and mention it's the meso form. Since it's a text response, I can draw using characters. Let's format nicely. For a) "represente a estrutura tridimensional de Y." I can provide a Fischer projection: ``` CH3 | CH2 | H — C — CH3 | H — C — CH3 | CH2 | CH3 ``` But this is not explicitly tridimensional? Fischer is a projection. Alternatively, use wedge/dash in a line drawing: CH3–CH2–C(H)(CH3)–C(H)(CH3)–CH2–CH3 with one C having CH3 wedge and H dash, the other having CH3 dash and H wedge. I can describe: "Y é o meso-3,4-dimetil-hexano. Estrutura tridimensional (projeção de Fischer): ..." Need ensure IUPAC name of X: 2-clorobutano. In Portuguese, "clorobutano" or "cloro-butano"? IUPAC: 2-clorobutano. The question says "cloreto de alquila (X)" so X is an alkyl chloride. IUPAC name: 2-clorobutano. Let's double-check if there is any alternative: What if X is 2-chlorobutane but the product is 3,4-dimethylhexane. But 3,4-dimethylhexane has two chiral centers. The meso form is optically inactive. The racemic form is also optically inactive if racemic. The question says "não é capaz de realizar desvio da luz polarizada, sendo esse composto (Y) com o menor número de carbonos possível nessas condições." It doesn't specify "meso" but "composto (Y)" singular. The smallest compound with two chiral carbons that is optically inactive as a single compound is meso-3,4-dimethylhexane. So yes. Let's also consider if Y could be 2,3-dichlorobutane? No, Wurtz of alkyl chloride gives alkane, not alkyl chloride. The product Y is hydrocarbon. Could the reaction be Wurtz with sodium metal and alkyl chloride, but "quantidade equivalente" maybe means 2 mol X with 2 mol Na, producing an alkane. Yes. Now, let's craft final answer in Portuguese. Include development. We need show reasoning: - Reação de Wurtz: 2 R-Cl + 2 Na → R-R + 2 NaCl. - Y tem 2 carbonos quirais e é opticamente inativo; menor número de carbonos → deve ser um alcano meso com dois carbonos quirais. - O menor alcano assim é o 3,4-dimetil-hexano (meso). - X é o cloreto de alquila que por acoplamento gera Y. Para Y = 3,4-dimetil-hexano, X = 2-clorobutano. - Estrutura tridimensional: projeção de Fischer do meso-3,4-dimetil-hexano. - Nomenclatura IUPAC de X: 2-clorobutano. Let's write final answer with the Fischer projection. I can use a code block or simple text. Fischer projection: ``` CH3 | CH2 | H - C - CH3 | H - C - CH3 | CH2 | CH3 ``` But this is not quite correct because the vertical chain should be continuous; the CH3 on right are substituents. The above is fine. However, in Fischer, the carbon chain is vertical, and the substituents are horizontal. At C3, top is C2 (CH2), bottom is C4. At C4, top is C3, bottom is C5 (CH2). So the drawing is correct. But the H and CH3 are on the same side for both? Yes, H left, CH3 right. That is meso. If both CH3 on right, then it's meso. If one CH3 right and one left, it's racemic. Wait, let's verify with the actual meso-3,4-dimethylhexane. I recall that meso-3,4-dimethylhexane has a plane of symmetry. In Fischer, if both methyl groups are on the same side, it's meso. If on opposite sides, it's racemic. So my drawing is correct. But note: In Fischer, the vertical bonds are going away. The main chain is vertical. At C3, the substituents are CH3 and H. At C4, CH3 and H. Both CH3 on right, H on left. This molecule has a plane of symmetry? If you draw a horizontal line between C3 and C4, the top half is CH3-CH2-C(H)(CH3) and bottom half is C(H)(CH3)-CH2-CH3. They are mirror images? Actually top half: C3 has H left, CH3 right. Bottom half: C4 has H left, CH3 right. They are identical, so there is a plane of symmetry perpendicular to the page? Yes, the molecule is meso. Now, the IUPAC name of X: 2-clorobutano. In Portuguese, it's common to write "2-clorobutano". Or "2-cloro-butano"? IUPAC rules: no hyphen between chloro and butane? Actually "2-clorobutano" is correct. Some might write "2-cloro-butano". I'll use "2-clorobutano". Let's also mention that X é o 2-clorobutano (racemic, pois deve gerar o meso). But the question just asks nomenclature, so no need. Final answer format: "Resposta final: a) [estrutura]; b) 2-clorobutano". But the instruction says: "Termine com uma linha neste formato: Resposta final: <o resultado de cada item, de forma curta>". So I need put the final answer at the end. The development can be before. Let's write in Portuguese. Need ensure the Fischer projection is clear. I'll include it in the development and then in final answer maybe short. Let's compose final response. Potential issue: The question says "represente a estrutura tridimensional de Y". In a text answer, a Fischer projection is acceptable. I'll label it as projeção de Fischer. Also, maybe they want the structure with wedges/dashes. I can also provide a wedge-dash description: "Y = meso-3,4-dimetil-hexano: CH3CH2C(H)(CH3)C(H)(CH3)CH2CH3, com um centro em R e outro em S." But "estrutura tridimensional" maybe they want a drawing. I'll provide Fischer. Let's double-check the number of carbons in 3,4-dimethylhexane: 6 main chain + 2 methyl = 8. Yes. Could there be a 7-carbon meso alkane with two chiral carbons? Let's try to construct: Need two chiral carbons. Each chiral carbon must have four different groups. The smallest groups are H, CH3, and something else. If the two chiral carbons are connected by a chain, the smallest total carbons? Consider 2,3-dimethylpentane: C2 attached to CH3, C3, CH3, H. Two CH3? But if C3 is chiral, the two CH3 on C2 are diastereotopic. However, is C2 considered a chiral center? In some definitions, a carbon with two identical groups is not a stereocenter even if they are diastereotopic. But in IUPAC, a stereocenter is an atom such that interchange of two ligands leads to a stereoisomer. For C2 in 2,3-dimethylpentane, if you interchange the two CH3 groups, do you get a different stereoisomer? Let's test: 2,3-dimethylpentane has configurations at C3 (R or S). At C2, there are two methyl groups: one is C1, one is branch. If you swap them, the molecule becomes a different diastereomer? Actually, consider (2R,3R)-2,3-dimethylpentane? But C2 has two CH3, so it cannot be R/S because two groups identical. However, if you label the two CH3 as pro-R and pro-S, swapping them might give the same molecule after rotation? Let's check with a model. 2,3-dimethylpentane: CH3-CH(CH3)-CH(CH3)-CH2-CH3. The two methyl groups on C2 are one attached to C2 and one attached to C2. They are constitutionally equivalent. But the molecule has a chiral center at C3. The two methyls on C2 are diastereotopic. If you replace one with ethyl, you get either 2-ethyl-3-methylpentane or 3-ethyl-2-methylpentane? Let's see: replace C1 methyl with ethyl: CH3CH2-CH(CH3)-CH(CH3)-CH2-CH3. That is 3,4-dimethylhexane? Actually chain: CH3CH2-C2H(CH3)-C3H(CH3)-CH2CH3. That is 3,4-dimethylhexane. Replace branch methyl with ethyl: CH3-CH(CH2CH3)-CH(CH3)-CH2-CH3. That is 3-ethyl-2-methylpentane? Longest chain: CH3-CH(CH2CH3)-CH(CH3)-CH2-CH3. The longest chain could be 6? Let's name: from ethyl end: CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3? Actually structure: CH3-CH(CH2CH3)-CH(CH3)-CH2-CH3. The longest chain: choose chain through ethyl and C2 and C3 and C4 and C5? That's 6 carbons: CH3-CH2-CH(CH3)-CH(CH3)-CH2-CH3? Wait, if you start from the ethyl end: CH3-CH2-C2H(CH3)-C3H(CH3)-CH2-CH3. That is exactly 3,4-dimethylhexane. So both replacements give 3,4-dimethylhexane! That means the two methyl groups on C2 are equivalent in terms of constitution, but if you replace one you get the same compound? Actually replacing C1 methyl or branch methyl both give 3,4-dimethylhexane? Let's check: Original 2,3-dimethylpentane: C1-C2-C3-C4-C5. C2 has methyl branch, C3 has methyl branch. Replace C1 methyl with ethyl: C1 becomes ethyl, so chain becomes C1-C2-C3-C4-C5 with ethyl at C1? That's 3,4-dimethylhexane? Let's number: CH3CH2-C2H(CH3)-C3H(CH3)-CH2-CH3. Longest chain: from ethyl end: C1' CH3, C2' CH2, C3' C2H(CH3), C4' C3H(CH3), C5' CH2, C6' CH3. That's 3,4-dimethylhexane. Replace branch methyl on C2 with ethyl: C2 becomes CH(CH2CH3)-CH(CH3)-CH2-CH3. Structure: CH3-CH(CH2CH3)-CH(CH3)-CH2-CH3. Longest chain: from the original C1: CH3-C2H(CH2CH3)-C3H(CH3)-CH2-CH3. To get longest chain, go through ethyl: CH3-CH2-C2H(CH3)-C3H(CH3)-CH2-CH3. Same 3,4-dimethylhexane. So the two methyls on C2 are homotopic? Actually they are equivalent because replacing either gives the same product. So C2 is not a stereocenter. Therefore 2,3-dimethylpentane has only one chiral center (C3). So no. Thus 8 carbons is minimum. Good. Now, final answer. Let's write in Portuguese: "Como a reação é de Wurtz, 2 R-Cl + 2 Na → R-R + 2 NaCl. O produto Y é um alcano. Para que Y tenha dois carbonos quirais e seja opticamente inativo, ele deve ser um composto meso. O menor alcano meso com dois carbonos quirais é o 3,4-dimetil-hexano (meso). Logo, R = 2-butila, e X = 2-clorobutano. a) Estrutura tridimensional (projeção de Fischer) do meso-3,4-dimetil-hexano: ... b) Nomenclatura IUPAC de X: 2-clorobutano." Final line: "Resposta final: a) meso-3,4-dimetil-hexano (projeção de Fischer: ...); b) 2-clorobutano." I'll include the Fischer in final answer. Let's ensure the Fischer is correctly formatted. Use spaces. ``` CH3 | CH2 | H — C — CH3 | H — C — CH3 | CH2 | CH3 ``` But note: In Fischer, the vertical chain should be straight. The above has CH2 and CH3 above and below. It's okay. Alternatively, I can draw a wedge-dash notation: CH3–CH2–C(H)(CH3)–C(H)(CH3)–CH2–CH3, with one center having CH3 wedge and H dash, the other having CH3 dash and H wedge. That is more explicitly tridimensional. I can write: "Y = meso-3,4-dimetil-hexano: CH3CH2C(H)(CH3)C(H)(CH3)CH2CH3, onde um centro é R e o outro S (ou, em cunha/traço, CH3 e H opostos em cada centro)." But Fischer is standard. I'll provide both description and Fischer. Now, final.

**Exploring Chiral Properties** I'm currently analyzing the reaction of an alkyl chloride with sodium metal. The key is understanding the structure of product Y, specifically its two chiral centers and its inability to rotate polarized light. This suggests a meso compound is formed. **Synthesizing Alkane Product** I'm focusing on the Wurtz reaction of an alkyl chloride with sodium. The goal is to identify product Y, an alkane with two chiral centers that is optically inactive, likely a meso compound. I'm now working to represent its 3D structure and determine the IUPAC name for the original alkyl chloride (X). **Determining Meso Compound Structure** I'm confirming that an optically inactive compound with chiral centers is indeed a meso compound due to internal compensation. My focus is on minimizing the carbon count in product Y, where the new carbon-carbon bond forms between two chiral centers. I'm now exploring the possible substituents on these chiral carbons to satisfy the meso condition. **Forming meso alkane** I'm investigating the Wurtz reaction where two identical radicals couple. To form product Y, a meso compound with two chiral centers, each central carbon must be attached to three distinct groups and the other coupled carbon. I'm currently minimizing carbon count to determine the initial alkyl chloride X and its structure. **Investigating Meso Alkane** I'm now considering scenarios where the chiral carbons are *not* the ones involved in the Wurtz coupling. My current hypothesis is to explore alkanes with fewer than 8 carbons, looking for the smallest acyclic alkane that can possess two chiral centers, thus potentially forming a meso compound. **Finding Smallest Chiral Alkanes** I've confirmed that 2,3-dimethylbutane is not chiral due to identical methyl groups on carbons 2 and 3. Linear hexane and 3-methylpentane also lack chiral centers. I've determined that 6-carbon acyclic alkanes cannot be chiral, and the smallest chiral alkane requires 7 carbons. **Identifying Alkane Substituents** I'm now focusing on the specific substituents required for two adjacent chiral carbons in an acyclic alkane. To achieve chirality at both, each carbon needs a hydrogen, the coupled carbon, and two distinct alkyl groups. This logic points towards an 8-carbon structure as the minimum to satisfy these requirements. **Confirming Eight-Carbon Structure** I've confirmed that the smallest acyclic alkane with two chiral centers requires eight carbons. My analysis indicates 3,4-dimethylhexane fits this criterion, as carbons 3 and 4 are indeed chiral and the molecule exhibits symmetry, allowing for a meso form. **Confirming Reaction Products** I've confirmed that meso-3,4-dimethylhexane (Y) is the smallest possible acyclic alkane with two chiral centers that is optically inactive, as stated in the problem. This means the original alkyl chloride (X) must be 2-chlorobutane, as its symmetrical coupling via the Wurtz reaction yields Y. **Confirming Smallest Meso Compound** I've confirmed that meso-3,4-dimethylhexane is indeed the smallest acyclic alkane that fits all the given criteria. This implies that the original alkyl chloride, X, must be 2-chlorobutane. I am now focusing on preparing the three-dimensional representations of Y as requested. **Visualizing Meso Stereochemistry** I'm now focusing on creating clear 3D representations of meso-3,4-dimethylhexane, using wedge-and-dash notation. I will also ensure the Fischer projection highlights the internal symmetry, confirming its meso nature. I am considering the R/S configurations on carbons 3 and 4. **Assigning R/S Configurations** I'm applying Cahn-Ingold-Prelog rules to assign R/S configurations to carbons 3 and 4 in meso-3,4-dimethylhexane. Based on priority assignments and the Fischer projection, I've determined carbon 3 is (3R) and carbon 4 is (4S), confirming the meso stereochemistry. I'm now translating this to wedge-and-dash notation. **Refining Wedge-and-Dash** I'm refining the wedge-and-dash representations, considering the zig-zag conformation and assigning substituents based on a plane. I'm determining the R/S configurations for carbons 3 and 4 by evaluating group priorities within this framework, focusing on the forward and backward projections. **Refining Wedge-and-Dash** I'm now focusing on the 3D representations. With the zig-zag conformation and a focus on the plane, I'm assigning priorities to determine R/S for carbons 3 and 4. I've found C3 is (3R). Now I'm confirming C4's configuration, considering that for a meso compound, the group in front at C3 should be behind at C4. **Confirming Chirality Configurations** I've successfully assigned the R/S configurations to both chiral carbons in meso-3,4-dimethylhexane as (3R) and (4S). This confirmed the molecule's meso nature and its center of inversion. I am now preparing the standard zig-zag perspective drawings and the Fischer projection to fully illustrate its stereochemistry. **Finalizando Diagramas Específicos** I'm now completing the precise wedge-and-dash and Fischer projections for meso-3,4-dimethylhexane. I'm meticulously ensuring the configurations (3R, 4S) are clearly depicted and that the internal symmetry of the meso compound is evident in each representation. **Confirmando a Menor Cadeia** I've confirmed that starting with 2-chloropropane yields 2,3-dimethylbutane. This molecule has two potential chiral centers but lacks optical activity due to the presence of two identical methyl groups on those carbons. This reinforces my conclusion about the minimum carbon requirements. **Confirmando Menor Alquila Cloreto** The prompt indicates that product Y has two chiral centers and is optically inactive, strongly suggesting a meso compound. My prior analysis confirmed that the smallest acyclic alkane meeting these criteria is 3,4-dimethylhexane. Given the Wurtz reaction couples two identical alkyl halide radicals, the original alkyl chloride (X) must therefore be 2-chlorobutane. My current focus is on reiterating this conclusion and its justification based on the number of carbons. **Reiterando o Cloreto Alquila** Confirmado que 2-clorobutano é o cloreto alquila de partida (X). Seu acoplamento via reação de Wurtz gera meso-3,4-dimetil-hexano (Y), o menor alcano acíclico com dois centros quirais e inatividade óptica, atendendo a todos os critérios.

**Analyzing Chemistry Problem** I am currently dissecting a chemistry problem from the IME 2026/2027 exam. My focus is on understanding the reaction between an alkyl chloride and sodium metal, and characterizing the resulting product. **Determining Alkane Structure** I'm working through the implications of Y being a meso compound with two chiral centers and the minimum possible carbon count. This suggests Y is a symmetrical alkane, and I'm focusing on how R must be structured to fulfill these criteria and lead to a meso product. **Exploring Dimerization Symmetry** I'm analyzing how R's structure dictates the chirality of the resulting dimer Y. I'm considering if chirality arises from R groups themselves or from the newly formed carbons, focusing on the minimum carbon count for a symmetric, meso dimer. **Testing Alkane Configurations** I'm evaluating the *sec*-butyl group as the R substituent. My analysis shows that this leads to 3,4-dimethylhexane, which has two chiral centers. I've confirmed this structure has the minimum number of carbons possible for a meso compound, precisely 8. **Confirming Meso Configuration** I'm confirming that $(3R,4S)$-3,4-dimethylhexane is indeed the meso isomer of Y. I'm also re-evaluating the chirality of X, 2-chlorobutane, and considering the implications of the Wurtz reaction on enantiomeric purity of the starting material. **Verifying Meso Stereochemistry** I'm now meticulously checking the R/S configurations of 3,4-dimethylhexane, specifically ensuring the $(3R,4S)$ designation accurately reflects the meso isomer. My plan is to draw both wedge-dash and Fischer projections to unequivocally demonstrate the plane of symmetry for this exam. **Confirming Meso Configuration** I am confirming that $(3S,4R)$-3,4-dimethylhexane is indeed the meso isomer of Y. I am also re-evaluating the chirality of X, 2-chlorobutane, and considering the implications of the Wurtz reaction on enantiomeric purity of the starting material. **Confirming Meso Symmetry** I am rigorously verifying the meso configuration by analyzing the 3D structure and relative positions of substituents. I'm paying close attention to how the zig-zag representation of the carbon chain influences the stereochemistry at C3 and C4, aiming to confirm the inverted orientation of key groups. **Visualizing Meso Stereochemistry** I am confirming the (3R,4S) configuration for 3,4-dimethylhexane by analyzing the zig-zag representation. I've established that the hydrogen must be on a wedge and the methyl on a dash at C4 for the S configuration, accurately depicting the meso isomer.