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Topic 17 Organic 2 Notes

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Total cards: 24•Content version: v3•Public deck•Updated: 8/1/2026

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#1
Front (Question)

No. 1 requirement for optical isomerism?

Back (Answer)

All 4 substituents on the central carbon must be different.

#2
Front (Question)

What name is given to the 2 stereoisomers of a carbon with 4 different substituents?

Back (Answer)

Enantiomers.

#3
Front (Question)

Enantiomers

Back (Answer)

Non-superimposable mirror images of each other.

#4
Front (Question)

Chiral centre

Back (Answer)

The carbon with 4 different groups attached. Often labelled with *. A molecule can have more than one chiral centre.

#5
Front (Question)

What is special about a chiral molecule?

Back (Answer)

It can rotate the plane of polarised light.

#6
Front (Question)

Optically active

Back (Answer)

Term given to chiral compounds because they rotate the plane of polarised light.

#7
Front (Question)

Achiral compounds

Back (Answer)

Optically inactive.

#8
Front (Question)

Why are racemic mixtures/racemates optically inactive?

Back (Answer)

For every molecule in the mixture that rotates the plane of polarisation in one direction, there is an enantiomer that rotates the plane in the opposite direction.

#9
Front (Question)

Why does light emerge from a racemic mixture with its plane of polarisation unchanged?

Back (Answer)

There is an equal concentration of both enantiomers, so the rotations cancel out.

#10
Front (Question)

Wedge-&-dash notation

Back (Answer)

The wedge comes towards you, out of the plane. The dashed line goes away from you, into the plane. The two straight lines are next to each other.

#11
Front (Question)

Why is SN1 unimolecular?

Back (Answer)

It has only one species in the rate equation.

#12
Front (Question)

Why is SN2 bimolecular?

Back (Answer)

It has 2 species in the rate equation.

#13
Front (Question)

Chloride nucleophile + butanol (secondary alcohol) by SN1

Back (Answer)

Cl - can approach the cation from either side as the molecule is planar. Thus, both enantiomers form; the race mate is not optically active.

#14
Front (Question)

Chloride ion + butanol (secondary alcohol) by SN2

Back (Answer)

Rear-side attack, the chloride attacks the partially positive chiral carbon. Inversion occurs, and the Cl switches with the OH group. Only one enantiomer is formed, so the product is optically active.

#15
Front (Question)

Carbonyls

Back (Answer)

Aldehydes & ketones.

#16
Front (Question)

-ate

Back (Answer)

Almost always 3 or 4 oxygens.

#17
Front (Question)

-ide

Back (Answer)

Just the element.

#18
Front (Question)

Break down the name: phosphide

Back (Answer)

Reduced form of phosphorous.

#19
Front (Question)

Break down the name: phosphate

Back (Answer)

Oxidised form of phosphorous.

#20
Front (Question)

How to find the name from the formula

Back (Answer)

Use oxidation states, and work backwards from the acid.

#21
Front (Question)

Why do tertiary molecules react via SN1?

Back (Answer)

It forms a more stable carbocation by the positive inductive effect.

#22
Front (Question)

Aldehyde functional group

Back (Answer)

R-CHO

#23
Front (Question)

Carbonyl functional group

Back (Answer)

R-COR’

#24
Front (Question)

Under carefully-controlled conditions, what are primary alcohols converted into?

Back (Answer)

Aldehydes.