The second question, I was less sure of, the amine can attack the carbonyl from

Question

The second question,

I was less sure of,

the amine can attack the carbonyl from the front or the back resulting in an enamine with the C-N bond pointing into the page or out at me,

these are not mirror images of each other,

and are therefore diastereomers?

are they optically active?

For this last question,

I put C, as the tertiary alcohol generated is achiral?

due to both sides of the ring to which it is attached being symmetrical.

Please explain if incorrect, thanks.

7. Which of the following statements describes the product(s) obtained in the following reaction? C) HO PTSA Product(s) Dean-Stark Trap, heat (A) The product will be a single enantiomer of an optically active acetal. (B) The product will be an achiral acetal. (C) The product will be a diastereomeric mixture of optically inactive acetals. (D) The product will be a diastereomeric mixture of optically active acetals. (E) The product will be an achiral hemiacetal. (F) The product will be racemic mixture composed of two hemiacetal enantiomers 8. Which of the following statements describes the product(s) obtained in the following reaction? :IN ) C) PTSA Product(s) Dean-Stark Trap, heat (A) The product will be a single enantiomer of an optically active imine. (B) The product will be an achiral enamine. (C) The product will be a racemic mixture of chiral enamines. (D) The product will be a diastereomeric mixture of optically active enamines. (E) A single enantiomerically pure enamine will be obtained in this reaction.

Explanation / Answer

7. Ans : B (The product is achiral due to presence of plane of symmetry).

8. Ans : B (The enamine is also achiral due to presence of plane of symmetry. Eneamine can generate both sides of carbonyl. Therefore enamine also have plane of symmetry).

9. Ans : C (The product is achiral due to plane of symmetry. NaBH4 will attacks from axial results equatorial alcohol).

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