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Organic analysis

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (24)

#1
Front (Question)

What is the reagent and positive test result for primary and secondary alcohols

Back (Answer)

Reagent: acidified pottasium dichromate (K₂Cr₂O₇) Primary and secondary alcohols are oxidised by K₂Cr₂O₇. Will turn from orange to green (Cr₂O₇²⁻ to Cr³⁺)

#2
Front (Question)

What is the reagent and positive test result for a tertiary alcohol

Back (Answer)

Will remain orange on addition of acidified potassium dichromate

#3
Front (Question)

What is the reagent and positive test result for alkenes

Back (Answer)

Reagent: Bromine water - goes from orange to colourless in the presence of an alkene

#4
Front (Question)

What is the reagent and positive test result for an aldehyde

Back (Answer)

Reagent: Fehling’s solution (in hot water bath) - goes from a blue solution to a red precipitate in the presence of an aldehyde or Reagent: Tollens reagent (in hot water bath)- formation of a silver mirror in the presence of an aldehyde

#5
Front (Question)

What is the reagent and positive test result for a carboxylic acid Write an equation for the positive test with CH3CO2H

Back (Answer)

Reagent: Add sodium carbonate Effervescence of CO2 formed - can then bubble through limewater to test this 2CH3CO2H + Na2CO3 —. 2CH3CO2Na + H2O + CO2

#6
Front (Question)

What is the reagent and positive test result for chloroalkanes

Back (Answer)

Reagent: Warm with silver nitrate Slow formation of white precipitate AgCl

#7
Front (Question)

Write an equation for the reaction of CH3CHO with tollens. Write the practical steps to test it and what occurs

Back (Answer)

Tollens’ Reagent formed by mixing aqueous ammonia and silver nitrate, which we heat gently in a water bath then add what were testing Aldehydes are oxidised by tollens into a carboxylic acid, and silver ions are reduced to silver atoms (silver mirror) CH3CHO + 2Ag+ + H2O —> CH3COOH + 2Ag + 2H+

#8
Front (Question)

Write an equation for the reaction of CH3CHO with Fehlings. Write the practical steps to test it and what occurs

Back (Answer)

We heat gently in a water bath then add what we are testing Aldehydes are oxidised by Fehling’s Solution into a carboxylic acid. The copper (II) ions are reduced to copper(I) oxide (blue to brick red) CH3CHO + 2Cu2+ + 2H2O —-> CH3COOH + Cu2O + 4H+

#9
Front (Question)

What is the m+1 or molecular ion peak

Back (Answer)

The peak with the highest mass/charge ratio which will be normally due to the original molecule that hasn’t fragmented, and therefore be equal to Mr

#10
Front (Question)

When is high resolution mass spectrometry useful

Back (Answer)

When identifying different molecules with he same Mr rounded to the nearest whole number

#11
Front (Question)

When does M+2 peak apply

Back (Answer)

If a compound contains a chlorine or a bromine atom then two molecular ion peaks will occur : a M and a M+2 peak will occur due to the two naturally occurring isotopes of chlorine or bromine.

#12
Front (Question)

What is the ratio of the 2 isotopes of Cl

Back (Answer)

Chlorine exists as Cl35 (75%) and Cl37 (25%) Bromine exists as Br79 (50%) and Br81 (50%)

#13
Front (Question)

What is the ratio of the heights of the M and M+2 peak in CH3Cl

Back (Answer)

CH3Cl will have a m/z value of M of 50 (CH3Cl 35) and M+2 of 52 (CH3Cl 37) The ratio of heights M:M+2 will be 3:1

#14
Front (Question)

What is the ratio of the heights of the M and M+2 peak in CH3Br

Back (Answer)

CH3Br will have m/z value of M of 94 (CH3Br 79) and M+2 of 96 (CH3Br 81) The ratio of heights M:M+2 will be 1:1

#15
Front (Question)

When will an M+4 peak occur

Back (Answer)

If a compound contains two chlorine or bromine atoms

#16
Front (Question)

What is the ratio of the heights of the M, M+2 and M+4 peak in C2H4Cl2

Back (Answer)

C2H4Cl2 will have a m/z value of M of 98 (C2H4Cl(35)Cl(35)) , a M+2 of 100 (C2H4Cl(35)Cl(37)) and a M+4 of 102 (C2H4Cl(35)Cl(37)) The ratio of heights M:M+2: M+4 will be 9:6:1

#17
Front (Question)

What is the ratio of the heights of the M, M+2 and M+4 peak in C2H4Br2

Back (Answer)

C2H4Br2 will have a m/z value of M of 186 (C2H4Br(79)Br(79)) , a M+2 of 188 (C2H4Br(79)Br(81)) and a M+4 of 190 (C2H4Br(81)Br(81)) The ratio of heights M:M+2: M+4 will be 1:2:1

#18
Front (Question)

What is the ratio of the heights of the M, M+2 and M+4 peak in C2H3Cl3

Back (Answer)

C2H3Cl3 will have a m/z value of M of 132 (C2H4Cl(35)Cl(35)Cl (35)) , a M+2 of 134 (C2H4Cl(35)Cl(35)Cl(37) , a M+4 of 136 (C2H4Cl(35)Cl(37)Cl(37)) and a M+6 of 138 (C2H4Cl(37)Cl(37)Cl(37)) The ratio of heights M:M+2:M+4:M+6 will be 27:27:9:1

#19
Front (Question)

What does the frequency of infrared radiation absorbed by a covalent bond depend on

Back (Answer)

The atoms on either side of the bond The position of the bond in the molecule

#20
Front (Question)

Where do we look to identify a functional group in infrared spectroscopy

Back (Answer)

Above 1500 cm-1

#21
Front (Question)

What does the fingerprint region in infrared spectrometry tell us

Back (Answer)

The size and position of the peaks are unique to a particular molecule, we can compare this region against a known library of spectra to identify a molecule

#22
Front (Question)

Where is the fingerprint region in infrared spectrometry

Back (Answer)

Between 500cm-1 and 1500cm-1

#23
Front (Question)

What do extra peaks in the fingerprint region of an infrared spectra indicate

Back (Answer)

Impurities in the sample

#24
Front (Question)

Explain how the greenhouse effect works

Back (Answer)

Electromagnetic radiation from the sun reaches the earth and is absorbed by land and sea. Some of this radiation is re-emitted as infrared Covalent bonds in greenhouse gasses absorb this radiation and re-emit this back towards earth (which warms the earth up as it cant escape)