A carbohydrate X having molecular weight 180 g mol-1 has one primary alcoholic group and four secondary alcoholic groups. It reacts with acetic anhydride to form pentaacetate. The molecular weight of pentaacetate formed is

Question

A carbohydrate X having molecular weight 180 g mol-1 has one primary alcoholic group and four secondary alcoholic groups. It reacts with acetic anhydride to form pentaacetate. The molecular weight of pentaacetate formed is (A) 180 (B) 120 (C) 390 (D) 42

Tardigrade Admin · Accepted Answer

The possible structure of X and reaction with
Molecular weight of

C H 3 CO -

group = 43, since ‘H’ is replaced by

- COC H_{3}

, therefore, increase in weight per

COC H_{3}

is 42.
Molecular weight of glucose pentaacetate

= 180 + 5 \times 42 = 180 + 210 = 390 g m o l^{- 1}

Q. A carbohydrate X having molecular weight 180gmol−1 has one primary alcoholic group and four secondary alcoholic groups. It reacts with acetic anhydride to form pentaacetate. The molecular weight of pentaacetate formed is

180

120

390

42

The possible structure of X and reaction with Molecular weight of CH3CO− group = 43, since ‘H’ is replaced by −COCH3​, therefore, increase in weight per COCH3​ is 42. Molecular weight of glucose pentaacetate =180+5×42=180+210=390gmol−1

Q. A carbohydrate X having molecular weight $180 g m o l^{- 1}$ has one primary alcoholic group and four secondary alcoholic groups. It reacts with acetic anhydride to form pentaacetate. The molecular weight of pentaacetate formed is

The possible structure of X and reaction with
Molecular weight of $C H 3 CO -$ group = 43, since ‘H’ is replaced by $- COC H_{3}$ , therefore, increase in weight per $COC H_{3}$ is 42.
Molecular weight of glucose pentaacetate $= 180 + 5 \times 42 = 180 + 210 = 390 g m o l^{- 1}$