JEE Advance - Chemistry (1986 - No. 23)

Write the IUPAC name of :
CH3CH2CH = CHCOOH
Pent-1-en-1-oic acid
2-Pentenoic acid
Pent-3-en-1-oic acid
3-Pentenoic acid
2-Penten-1-ol

Explanation

IIT-JEE 1986 Chemistry - Basics of Organic Chemistry Question 41 English Explanation 1

Note :

(1) Here functional group $$-$$COOH present. All functional groups add as secondary suffix.

(a) If carbon atom of $$-$$COOH group present in main chain then secondary suffix name is $$\to$$ oic acid.

(b) If carbon atom of $$-$$COOH group present not in main chain then secondary suffix name is $$\to$$ carboxylic acid.

As here carbon atom of $$-$$COOH group present in main chain that is why we use oic acid.

(2) Functional group should always get lowest possible number.

(3) Alkane, alkene(=) and alkyne($$\equiv$$) always act as primary suffix.

For alkane we use word "ane".

For alkene we use word "ene".

For alkyne we use word "yne".

Here alkene(=) present so we use word "ene" as primary suffix.

(4) After last e of "ene" if we get a, i, o, u, y as the first letter of secondary suffix then we delete e.

Here in the name $$\to$$

IIT-JEE 1986 Chemistry - Basics of Organic Chemistry Question 41 English Explanation 2

(5) Naming is done by following way $$\to$$

Prefix + word root + suffix(p) + suffix(s)

(a) Here no prefix (name of side chain) as there is no side chain present.

(b) Word root represent number of carbon atom in main chain.

Here in main chain 5 carbon atoms present so word root is "pent".

(c) Primary suffix is alkene.

(d) Secondary suffix is function group $$-$$COOH. It's name oic acid.

(6) If functional group's carbon get number "1" then we can ignore it in nomenclature. So it's nomenclature can be also $$\to$$

$$\mathrm{pent - 2 - enoic\,acid}$$

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