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Chapter 12- Organic Chemistry- Some Basic Principles and Techniques Interview Questions Answers

Question 1 :

What are hybridisationstates of each carbon atom in the following compounds?

CH2=C=O, CH3CH=CH2,(CH3)2CO, CH2=CHCN, C6H6

Answer 1 :

(i) 

C–1 is sp2 hybridised.

C–2 is sp hybridised.

(ii) 

C–1 is sp3 hybridised.

C–2 is sp2 hybridised.

C–3 is sp2 hybridised.

(iii)

C–1 and C–3 are sp3 hybridised.

C–2 is sp2 hybridised.

(iv) 

C–1 is sp2 hybridised.

C–2 is sp2 hybridised.

C–3 is sp hybridised.

(v) C6H6

All the 6 carbonatoms in benzene are sp2 hybridised.

Question 2 :

Indicate the σ and π bondsin the following molecules:

C6H6,C6H12, CH2Cl2, CH=C = CH2, CH3NO2, HCONHCH3

Answer 2 :

(i) C6H6

There are six C–C sigma () bonds, six C–H sigma () bonds, and three C=C pi () resonating bonds inthe given compound.

(ii) C6H12

There are six C–C sigma () bonds and twelve C–H sigma () bonds in the givencompound.

(iii) CH2Cl2
There two C–H sigma () bonds and two C–Cl sigma () bonds in the givencompound.

(iv) CH=C = CH2

There are two C–C sigma () bonds, four C–H sigma () bonds, and two C=C pi () bonds in the givencompound.

(v) CH3NO2.
There are three C–H sigma () bonds, one C–N sigma () bond, one N–O sigma () bond, and one N=O pi () bond in the givencompound.
(vi) HCONHCH3
There are two C–N sigma () bonds, four C–H sigma () bonds, one N–H sigma bond, and one C=O pi () bond in the givencompound.

Question 3 :

Write bond line formulasfor: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.

Answer 3 :

The bond line formulaeof the given compounds are:

(a) Isopropyl alcohol

(b) 2, 3–dimethylbutanal

(c) Heptan–4–one

Question 4 :

Give the IUPAC names ofthe following compounds:

(a)

(b)

(c)

(d)

(e)

(f) Cl2CHCH2OH

Answer 4 :

(a)

1–phenyl propane

(b)

3–methylpentanenitrile

(c)

2, 5–dimethyl heptane

(d)

3–bromo–3–chloroheptane

(e)

3–chloropropanal

(f) Cl2CHCH2OH

2,2- Dichloroethanol

Question 5 :

Which of the followingrepresents the correct IUPAC name for the compounds concerned? (a)2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane(d) But-3-yn-1-ol or But-4-ol-1-yne

Answer 5 :

(a) The prefix di inthe IUPAC name indicates that two identical substituent groups are present inthe parent chain. Since two methyl groups are present in the C–2 of the parentchain of the given compound, the correct IPUAC name of the given compound is 2,2–dimethylpentane.

(b) Locant number 2,4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given compound is 2,4, 7–trimethyloctane.

(c) If thesubstituents are present in the equivalent position of the parent chain, thenthe lower number is given to the one that comes first in the name according tothe alphabetical order. Hence, the correct IUPAC name of the given compound is2–chloro–4–methylpentane.

(d) Two functionalgroups – alcoholic and alkyne – are present in the given compound. Theprincipal functional group is the alcoholic group. Hence, the parent chain willbe suffixed with ol. The alkyne group is present in the C–3 of theparent chain. Hence, the correct IUPAC name of the given compound isBut–3–yn–1–ol.

Question 6 :

Draw formulas for thefirst five members of each homologous series beginning with the followingcompounds. (a) H–COOH (b) CH3COCH3  (c) H–CH=CH2

Answer 6 :

The first five membersof each homologous series beginning with the given compounds are shown asfollows:

(a)

H–COOH : Methanoic acid

CH3–COOH :Ethanoic acid

CH3–CH2–COOH: Propanoic acid

CH3–CH2–CH2–COOH: Butanoic acid

CH3–CH2–CH2–CH2–COOH: Pentanoic acid

(b)

CH3COCH3 :Propanone

CH3COCH2CH:Butanone

CH3COCH2CH2CH:Pentan-2-one

CH3COCH2CH2CH2CH:Hexan-2-one

CH3COCH2CH2CH2CH2CH:Heptan-2-one

(c)

H–CH=CH2 :Ethene

CH3–CH=CH:Propene

CH3–CH2–CH=CH2 :1-Butene

CH3–CH2–CH2–CH=CH:1-Pentene

CH3–CH2–CH2–CH2–CH=CH2 :1-Hexene

Question 7 :

Give condensed and bondline structural formulas and identify the functional group(s) present, if any,for :

(a) 2,2,4-Trimethylpentane

(b)2-Hydroxy-1,2,3-propanetricarboxylic acid

(c) Hexanedial

Answer 7 :

(a) 2, 2, 4–trimethylpentane

Condensed formula:

(CH3)2CHCH2C (CH3)3

Bond line formula:

(b) 2–hydroxy–1, 2,3–propanetricarboxylic acid

Condensed Formula:

(COOH)CH2C(OH)(COOH)CH2(COOH)

Bond line formula:

The functional groupspresent in the given compound are carboxylic acid (–COOH) and alcoholic (–OH)groups.

(c) Hexanedial

Condensed Formula:

(CHO) (CH2)4 (CHO)

Bond line Formula:

The functional grouppresent in the given compound is aldehyde
(–CHO).

Question 8 :

Identify the functionalgroups in the following compounds

(a) 

(b) 

(c)

Answer 8 :

The functional groupspresent in the given compounds are:

(a) Aldehyde (–CHO),

Hydroxyl(–OH), 

Methoxy (–OMe),

C=C double bond  

(b) Amino (–NH2);primary amine,

Ester (-O-CO-), 

Triethylamine (N(C2H5)2);tertiary amine

(c) Nitro (–NO2),

C=C double bond 

Question 9 :

Which of the two: O2NCH2CH2O orCH3CH2O is expected to be more stableand why?

Answer 9 :

NO2 groupis an electron-withdrawing group. Hence, it shows –I effect. By withdrawing theelectrons toward it, the NO2 group decreases thenegative charge on the compound, thereby stabilising it. On the other hand,ethyl group is an electron-releasing group. Hence, the ethyl group shows +Ieffect. This increases the negative charge on the compound, therebydestabilising it. Hence, O2NCH2CH2O isexpected to be more stable than CH3CH2O.

Question 10 :

Explain why alkyl groupsact as electron donors when attached to a π system.

Answer 10 :

When an alkyl groupis attached to a π system, it acts as an electron-donor group by theprocess of hyperconjugation. To understand this concept better, let us take theexample of propene.

In hyperconjugation, thesigma electrons of the C–H bond of an alkyl group are delocalised. This groupis directly attached to an atom of an unsaturated system. The delocalisationoccurs because of a partial overlap of a sp3 –s sigmabond orbital with an empty orbital of the π bondof an adjacent carbon atom.

The process ofhyperconjugation in propene is shown as follows:

This type of overlapleads to a delocalisation (also known as no-bond resonance) ofthe π electrons, making the molecule more stable.


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Chapter 12- Organic Chemistry- Some Basic Principles and Techniques Contributors

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