AQA A level Chem 3.1 Introduction to Organic Chemistry

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Last updated 2:47 PM on 3/14/26
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4 Terms

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Define the term stereoisomers. (2)

(Different molecules/compounds with the) same (molecular and) structural formula (1)

Different spatial arrangement of atoms (1)

2
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Under suitable conditions, 2-bromobutane reacts with sodium hydroxide to

produce a mixture of five products, A, B, C, D and E.

Products A, B and C are alkenes.

A is a structural isomer of B and C.

A does not exhibit stereoisomerism.

B and C are a pair of stereoisomers.

Products D and E are alcohols.

D and E are a pair of enantiomers.

Deduce the name of isomer A. Explain why A does not exhibit stereoisomerism. (2)

A = but-1-ene (1)

two groups/atoms/Hs the same on one of the C=C carbons (1)

3
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Under suitable conditions, 2-bromobutane reacts with sodium hydroxide to

produce a mixture of five products, A, B, C, D and E.

Products A, B and C are alkenes.

A is a structural isomer of B and C.

A does not exhibit stereoisomerism.

B and C are a pair of stereoisomers.

Products D and E are alcohols.

D and E are a pair of enantiomers.

Deduce the name of isomer B and the name of isomer C. Explain the origin of the stereoisomerism in B and C. (2)

Z-but-2-ene AND E-but-2-ene (1)
lack of/restricted/no (free) rotation around C=C/double bond (1)

4
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<p>Compound J is one of a pair of stereoisomers</p><p>Explain:</p><p>• what stereoisomers are</p><p>• how E-Z stereoisomerism occurs</p><p>• how the Cahn-Ingold-Prelog rules can be used to decide whether compound J is an E or Z isomer. (6) </p>

Compound J is one of a pair of stereoisomers

Explain:

• what stereoisomers are

• how E-Z stereoisomerism occurs

• how the Cahn-Ingold-Prelog rules can be used to decide whether compound J is an E or Z isomer. (6)

Stereoisomers are compounds with the same structural formula but a different arrangement of atoms in space.

E/Z isomerism arises due to restricted rotation around the C=C bond

When each C of C=C is attached to two different groups

CH3CH2 is higher priority than H on RHS C

as atomic number of C is higher than H or C = 6 v H = 1

(CH3)2CH is higher priority than CH3CH2 on LHS C

as atomic numbers of atoms joined to C (joined to C=C) are higher for (CH3)2CH than CH3CH2

highest priority groups on opposite sides (of C=C)

Therefore compound J is a Z isomer (6)