In polysaccharides the linkage connecting monosaccharide units is… - Vidyadip

MCQ

In polysaccharides the linkage connecting monosaccharide units is called

✓
Glycoside linkage
B
Nucleoside linkage
C
Glycogen linkage
D
Peptide linkage

✓

Answer

Correct option: A.

Glycoside linkage

a
(a) That is called glycosidic linkage.

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The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy $(BDE)$ or Bond Strength.
$\text{BDE}$ is affected by $s-$character of the bond and the stability of the radicals formed.
Shorter bonds are typically stronger bonds. $\text{BDEs}$ for some bonds are given below
:$\begin{array}{l} Cl - Cl ( g ) \longrightarrow Cl ^*( g )+ Cl ^*( g ) \Delta H ^{\circ}=58 \ kcal \ mol ^{-1} \end{array}$
$ H _3 C - Cl ( g ) \longrightarrow H _3 C ^*( g )+ Cl ^{\circ}( g ) \Delta H ^{\circ}=85 \ kcal \ mol ^{-1} $
$ H - Cl ( g ) \quad \longrightarrow H ^*( g ) \quad+ Cl ^*( g ) \Delta H ^{\circ}=103 \ kcal \ mol ^{-1} $
$(1)$ Correct match of the $C - H$ bonds (shown in bold) in Column $J$ with their $\text{BDE}$ in Column $K$ is

Column $J$ Molecule	Column $K$ $\operatorname{BDE}( kcal mol -1)$
$(P)$ $H - C H \left( CH _3\right)_2$	${ (i) } 132$
$(Q)$ $H - CH _2 Ph$	${ (ii) } 110$
$(R)$ $H - C H = CH _2$	${ (iii) } 95$
$(S)$ $H - C \equiv CH$	${ (iv) } 88$

$(A) P - iii, Q - iv, R - ii, S - i$
$(B) P - i, Q - ii, R - iii, S - iv$
$(C) P - iii, Q - ii, R - i, S - iv$
$(D) P - ii, Q - i, R - iv, S - iii$
$(2)$ For the following reaction
$CH _4( g )+ Cl _2( g ) \xrightarrow{\text { light }} CH _3 Cl ( g )+ HCl ( g )$
the correct statement is
$(A)$ Initiation step is exothermic with $\Delta H ^{\circ}=-58 \ kcal \ mol ^{-1}$
$(B)$ Propagation step involving ${ }^{\circ} CH _3$ formation is exothermic with $\Delta H ^{\circ}=-2 \ kcal \ mol ^{-1}$.
$(C)$ Propagation step involving $CH _3 Cl$ formation is endothermic with $\Delta H ^{\circ}=+27 \ kcal \ mol ^{-1}$.
$(D)$ The reaction is exothermic with $\Delta H ^{\circ}=-25 \ kcal \ mol ^{-1}$.

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