5 Marks Questions

Question 15 Marks

(a) What type of protein is present in keratin?
(b) Write the reactions showing the presence of following in the open structure of glucose:
i. an aldehyde group
ii. a primary alcohol
(c) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
(d) Amino acids show amphoteric behaviour. Why?
(e) What are a Amino Acids? Give examples.
(f) a. How can you explain the absence of an aldehyde group in the pentaacetate of D-glucose?
b. Name the bases present in RNA. Which one of these is not present in DNA?
(g) Which monosaccharide units are present in starch, cellulose and glycogen and which linkages link these units?

Answer

(i) Fibrous Proteins
(ii)

(iii) When a nucleotide from the DNA containing thymine is hydrolyzed, thymine $\beta$-D-2-deoxyribose and phosphoric acid are obtained as products.
(iv) Amino acids contain both amino $\left(- NH _2\right)$ and carboxyl $(- COOH )$ groups, thus they react with both acids and bases. Hence, amino acids are amphoteric in nature.
(v) Those amino acids in which $- NH _2$ group and - COOH group are attached to same carbon are called $\alpha$-amino acids. These are obtained by hydrolysis of proteins. e.g., glycine.
(vi) a. The pentaacetate of glucose does not react with hydroxylamine / HCN / Schiff's reagent indicating the absence of free - CHO group. b. Adenine, Guanine, Uracil and Cytosine Uracil
(vii) n starch $\alpha$-glucose units are present, in cellulose $\beta$-D glucose units are present. In starch and glycogen glycosidic $\alpha$ linkage is present between $C 1- C 4$ and in cellulose glycosidic $\beta$-linkage is present between glucose units.

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Question 25 Marks

i. Give reasons :
a. Although $- \ce{NH2}$ group is o/p directing in electrophilic substitution reactions, yet aniline, on nitration gives good yield of $m-$nitroaniline.
b. $(\ce{CH 3)2NH}$ is more basic than $(\ce{CH3)3N}$ in an aqueous solution.
c. Ammonolysis of alkyl halides is not a good method to prepare pure primary amines.
ii. Distinguish between the following:
a. $\ce{CH 3 CH 2 NH 2}$ and $(\ce{CH 3 CH 2)2NH}$
b. Aniline and $\ce{CH3NH2}$

Answer

i. Nitration is carried out in acidic medium. In an acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, a substantial amount of meta derivative $(m-$nitroaniline$)$ is also formed.
b. $(\ce{CH3)2NH}$ is a secondary amine and $\ce{(CH3)3N}$ is a tertiary amine. Tertiary amine due to the presence of three alkyl groups is more hindered than secondary amine which has only two alkyl groups attached to it. Therefore formation of ammonium ion is easier in secondary amine than the tertiary amine. Therefore, it makes secondary amine less basic than the tertiary amine.
c. The ammonolysis of alkyl halide leads to the formation of the mixture of primary, secondary and tertiary amine along with the formation of quaternary salt. It is very difficult to separate pure primary amine from this mixture.
$(a)$

Test	$\ce{CH3CH2NH2}$	$(\ce{CH3CH2)2NΗ}$
Carbylamine test $($add chloroform and alcoholic $\ce{KOH}$ to both the compounds separately in a test tube$)$	Forms a foul$-$smelling compound $($gives positive test$)$	No reaction take place $($gives negative test$)$

$(b)$

Azo dye Test	Aniline	Methyl Amine $(\ce{CH3NH2})$
Add a small amount of nitrous acid with aq. $\ce{HCI}$	Forms a yellow coloured dye $($gives positive test$)$	No dye is formed $($gives negative test$)$

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Question 35 Marks

I. Show how p-aminoazobenzene can be obtained from aniline.
II. Write structures for the following compounds:
a. Benzene diazonium chloride
b. p-Nitrotoluene
c. Sulphanilic acid

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Question 45 Marks

Draw the structures of optical isomers of:
i. $\left[ Cr \left( C _2 O _4\right)_3\right]^{3-}$
ii. $\left[ PtCl _2( en )_2\right]^{2+}$
iii. $\left[ Cr \left( NH _3\right)_2 Cl _2( en )\right]^{+}$

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Question 55 Marks

Explain on the basis of valence bond theory that $\left[ Ni ( CN )_4\right]^{2-}$ ion with square planar structure is diamagnetic and the $\left[ NiCl _4\right]^{2-}$ ion with tetrahedral geometry is paramagnetic.

Answer

$Ni$ is in the $+2$ oxidation state i.e., in $d^8$ configuration.

There are $4 CN^-$ ions. Thus, it can either have a tetrahedral geometry or square planar geometry.
Since $CN^-$ ion is a strong field ligand, it causes the pairing of unpaired $3d$ electrons.

It now undergoes $dsp^2$ hybridization. Since all electrons are paired, it is diamagnetic. In case of $[NiCl_4]^{2-}, CN^-$ ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired $3d$ electrons.
Therefore, it undergoes $sp^3$ hybridization.

Since there are $2$ unpaired electrons in this case, it is paramagnetic in nature.

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Chemistry 5 Marks Questions

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