Aldehydes, Ketones and Carboxylic Acids · Question 37

[figure] 0.1 mole of compound 'S' will weigh g. (Given molar mass in g mol-1 C : 12, H : 1, O : 16)

Answer

13

Solution

Step 1: CrO3 (excess) oxidises both -OH groups: the 2 alcohol to a ketone and the 1 alcohol to an aldehyde (as per NCERT, CrO3 in anhydrous medium stops at the aldehyde).

[figure]

Step 2: One mole of ethylene glycol / H protects only the more reactive aldehyde group as a cyclic acetal (Q).

[figure]

Step 3: CH3MgI adds to the free keto group; H3O gives the 3 alcohol and also hydrolyses the acetal back to -CHO (R).

Step 4: NaBH4 reduces -CHO to -CH2OH (S).

[figure]

S = 2-(hydroxymethyl)-1-methylcyclopentan-1-ol, C7H14O2

Molar mass = 7 12 + 14 1 + 2 16 = 130 g mol-1

Mass of 0.1 mol = 0.1 130 = 13 g

Answer: 13