This question is about isomers with the molecular formula C₆H₁₄O - AQA - A-Level Chemistry - Question 6 - 2021 - Paper 2

To distinguish between the two enantiomers of the branched chain aldehyde C₆H₁₄O, you can use plane-polarized light. One enantiomer will rotate the plane of polarized light in a clockwise direction (dextrorotatory), while the other will rotate it in a counterclockwise direction (levorotatory).

The E and Z forms of a structural isomer of C₆H₁₄O can be represented as follows:

E Isomer

CH₃
 |
C=C–CHO
 |   
CH₂–C₂H₅

Z Isomer

CH₃
 |
C=C–CHO
 |   
C₂H₅–CH₂

These structures demonstrate both optical activity (due to the chiral center) and geometric isomerism (due to the C=C double bond).

Two cyclic isomers of C₆H₁₄O with an ether functional group are:

Isomer 1

    O
   / \
  C₅H₁₁–C

Isomer 2

   O
   |
C₄H₉–C
   / \
C₃H₇

Both of these structures have ether functional groups (C–O–C) and can be expected to show only three peaks in their ¹³C NMR spectra.