Question

The number of asymmetric carbon atoms in the glucose molecule in open and cyclic form is ______.

Options

• Four, Five

• Four, Four

• Five, Four

• Five, six

Answer 1

The number of asymmetric carbon atoms in the glucose molecule in open and cyclic form is Four, Five.

Explanation:

\[\begin{array}{cc}
\phantom{}\ce{CHO}\\
\phantom{}|\phantom{....}\\
\ce{H-C^*-OH}\phantom{.}\\
|\phantom{....}\\
\ce{HO-C^*-H}\phantom{.....}\\
|\phantom{....}\\
\ce{H-C^*-OH}\phantom{..}\\
|\phantom{....}\\
\ce{H-C^*-OH}\phantom{..}\\
|\phantom{....}\\
\phantom{...}\ce{\underset{D-(+)-Glucose}{CH2OH}}
\end{array}\]

α–D–(+)–Glucopyranose
(Haworth structure)

∴ Four asymmetric carbon, Five asymmetric carbon.