Organic Chemistry 9th Edition

Published by Brooks Cole
ISBN 10: 1305080483
ISBN 13: 978-1-30508-048-5

Chapter 7 - Alkenes: Structure and Reactivity - Problem: 3


Diazepam has 13 H atoms and a molecular formula of $C_{16}H_{13}ClN_{2}O$

Work Step by Step

The degree of unsaturation in a molecule is the number of multiple bonds and/or rings that the molecule possesses. Here, Diazepam has a total of 11 degrees of unsaturation (3 rings and 8 double bonds). This corresponds to the absence of 11 pairs of H atoms from the formula of the corresponding saturated hydrocarbon. Let there be $'x'$ number of H atoms in the formula, i.e. $C_{16}H_{x}ClN_{2}O$. The formula of its corresponding hydrocarbon is achieved by (i) Adding the number of halogen atoms to the number of H atoms. (ii) Subtracting the nmber of N atoms from the number of H atoms (iii) Ignoring the number of O atoms. The final number of H atoms will be: $x+1-2$ = $x-1$ And the formula of the hydrocarbon will be: $C_{16}H_{x-1}$ The saturated hydrocarbon with 16 carbons has the formula: $C_{16}H_{34}$ Now, since the molecule has 11 degrees of unsaturation; $H_{34} - H_{x-1} = H_{22} = 11 H_{2}$ i.e. (Number of H atoms in saturated hydrocarbon $-$ Number of H atoms in unsaturated hydrocarbon = Number of Pairs of H atoms missing from unsaturated hydrocarbon/Degree of unsaturation) which when simplified, gives us: $34-(x-1)=22$ $(x-1) = 34-22$ $(x-1) = 12$ $x = 13$ Therefore, the number of H atoms in Diazepam is 13.
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