Given:From the following compound(s), which on reaction with $H N O_{3}$ will give the product having degree of rotation, $[α]_{D} = - {52.7}^{\circ}$ is/are:

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Solution

The correct option is D

Since, we have to get the product $(x)$ of $[α]_{D} = - {52.7}^{\circ}$ i.e. the enantiomer of product (P) which is only possible from (c) & (d).
From (c), enantiomer of saccharic acid is obtained.
From (d), enantiomer of saccharic acid is obtained.

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Similar questions

H N O_{3}

[α]_{D} = - {52.7}^{\circ}

H N O_{3}

C O_{2}

N a H C O_{3}

α -

B r_{2} / h ν

3^{\circ}

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