An organic compound A (C7H16O) shows both enantiomerism and diastereomerism. Treatment of a pure enantiomer of A with Na2CrO4 /Dil. H2SO4 gives B(C7H14O) - Also A on dehydration with concentrated H2SO4 gives a single alkene C (C7H14). Ozonolysis of C followed by work-up with Zn-H2Ogives D (C5H10O) as one of the product which gives racemic mixture on reduction with NaBH4.Q.If B is reduced with LiAIH4 followed by acid hydrolysis will givea)a pure enantiomerb)a racemic mixturec)a pair of diastereomersd)an achiral alcoholCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Explanation:

Enantiomerism:
Enantiomerism refers to the phenomenon where two molecules are mirror images of each other and are non-superimposable. In this case, compound A shows enantiomerism, which means it exists in two different mirror-image forms.

Diastereomerism:
Diastereomerism refers to the phenomenon where two molecules have the same connectivity but are not mirror images of each other and are non-superimposable. In this case, compound A shows diastereomerism, which means it exists in multiple forms that are not mirror images of each other.

Reaction 1: Treatment of A with Na2CrO4/Dil. H2SO4:
When compound A is treated with Na2CrO4 and dilute H2SO4, it undergoes oxidation reaction and gives compound B. The reaction is as follows:

C7H16O (A) + Na2CrO4 / Dil. H2SO4 → C7H14O (B)

Reaction 2: Dehydration of A with concentrated H2SO4:
When compound A is dehydrated with concentrated H2SO4, it undergoes elimination reaction and gives a single alkene C. The reaction is as follows:

C7H16O (A) + H2SO4 (conc.) → C7H14 (C)

Reaction 3: Ozonolysis of C followed by work-up with Zn-H2O:
When alkene C is subjected to ozonolysis followed by treatment with Zn-H2O, it undergoes oxidative cleavage and gives compound D as one of the products. The reaction is as follows:

C7H14 (C) + O3 → C5H10O (D)

Reaction 4: Reduction of D with NaBH4:
When compound D is reduced with NaBH4, it undergoes reduction reaction and gives a racemic mixture. This means that both enantiomers of D are formed in equal amounts. The reaction is as follows:

C5H10O (D) + NaBH4 → Racemic mixture of D

Reaction 5: Reduction of B with LiAlH4 followed by acid hydrolysis:
When compound B is reduced with LiAlH4 followed by acid hydrolysis, it undergoes reduction reaction and gives a pair of diastereomers. This means that two different diastereomers of B are formed. The reaction is as follows:

C7H14O (B) + LiAlH4 → Diastereomers of B

Therefore, the correct answer is option c) a pair of diastereomers.

Answer

Since, A is completely saturated, it must contain more than one chiral carbon atoms in order to show both enantiomerism and diastereomerism. Also, C on ozonolysis gives D (C₅H₁₀O) as one product, other product must be CH₃SHO. Hence, C must be

B is enantiomeric. If a pure enantiomer of B is reduced with LiAIH₄, pair of diastereomers would be formed.

Answer

Since, A is completely saturated, it must contain more than one chiral carbon atoms in order to show both enantiomerism and diastereomerism. Also, C on ozonolysis gives D (C₅H₁₀O) as one product, other product must be CH₃SHO. Hence, C must be

B is enantiomeric. If a pure enantiomer of B is reduced with LiAIH₄, pair of diastereomers would be formed.

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