Things to note - when answering questions on topic 7 and 19:

0.0(0)
studied byStudied by 0 people
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/7

flashcard set

Earn XP

Description and Tags

Chemistry

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

8 Terms

1
New cards

Give one advantage and one disadvantage of using a chemical test rather than infrared spectroscopy to determine whether any of the 2-methylpropan-2-ol remained.

Advantages:

  • cost associated with the chemical test is small and the result is very rapid

Disadvantage:

  • chemical test is not so sensitive / uses some of the sample which cannot easily be recovered

2
New cards

State why HOOCCH=CHOOH has E/Z isomers

  • restricted / limited rotation about the C=C double bond

  • each carbon atom on the double bond is attached to two different atoms/ different groups (of atoms)/ to a (H) atom and a COOH group

3
New cards

The relative molecular mass of 2-methylpropan-2-ol is 74. Give a possible reason why there is no molecular ion peak in the mass spectrum of 2-methylpropan-2-ol.

Molecular ion/ molecule fragments/ is unstable

4
New cards

TMS is an inert and non-toxic compound. State two other reasons why TMS is suitable for use as a standard when recording NMR spectra.

  • single peak / all H or all C in same environment

  • peak to the right / upfield

  • low boiling temperature / volatile

  • gives a strong signal so only small amount needed

5
New cards

This question is about some carbonyl compounds with the molecular formula C5H10O. An aldehyde with molecular formula C5H10O has a 13C NMR spectrum with three peaks. The high resolution NMR spectrum of this aldehyde has two peaks and neither of them is split. Deduce the displayed formula of this aldehyde. Justify your answer by referring to both NMR spectra.

  • displayed formula of aldehyde (1)

  • three different carbon environments indicated (1)

  • two different proton environments indicated (1)

  • no splitting as there are no hydrogens on the adjacent carbon atom(s)

<ul><li><p>displayed formula of aldehyde (1)</p></li><li><p>three different carbon environments indicated (1)</p></li><li><p>two different proton environments indicated (1)</p></li><li><p>no splitting as there are no hydrogens on the adjacent carbon atom(s)</p></li></ul><p></p>
6
New cards

Explain why different compounds will have different retention times in the same column, under the same conditions.

  • retention time depends on the polarity or attraction/ affinity/ solubility/ of component for the stationary phase

  • the greater the attraction/ affinity/ solubility/ of the component for the stationary phase the greater the retention time

OR

  • retention time depends on boiling temperature of the compound

  • higher boiling temperature compounds spend less time in the gas phase/ mobile phase so have longer retention time

7
New cards

State why nitrogen and argon are suitable carrier gases for gas chromatography.

gases are inert and do not react with components of the mixture

8
New cards

The relative abundance of the abundance of the isotope 35Cl is 75.5%. The relative peak height of the isotope is 82.5 in the mass spectrum. Calculate the relative peak height of the missing peak caused by the isotope 37Cl.

knowt flashcard image