Answer:
The substance likely to form an insoluble precipitate is
Explanation:
The test for cation when is mixed in aqueous solution such as NaOH (sodium hydroxide) produces an insoluble precipitate
Answer:
The compound elucidated from the spectral data is <u>4-methyl penta-2-none</u>
Explanation:
- <u>1700 cm-1 from IR data</u> suspects aldehyde/ketone or carboxylic acid. However,since the peak is not a stretched vibration, it implies an aldehyde or ketone present.
- <u>3 ppm H NMR</u> confirms O-CH3 bond
- <u>24.4 13C NMR</u> confirms CH3-R bond
- 26.4 13C NMR confirms H3C-(R)C(H)-R
- 44.2 13C NMR Confirms C=
- 2126 13C NMR confirms aldehyde C=O bond
<u>The deduced structure is 4-methyl penta-2-one (see attached) </u>given multiple CH3 atoms.
That is approximately .63620 grams.
This is formula to find pH. -log[H^+]
now press calculator -log[4.7×10^11] and you will get 10.33
Answer:
B. A tertiary alcohol; a secondary alcohol
Explanation:
Grignard reagent has a formula RMgX where X is a halogen, and R is an alkyl or aryl group.
Grignard reagent reacts with aldehydes and ketones to form alcohol. However the classification (that is; Primary, secondary and tertiary) varies depending on which it reacts with.
2-butanone is a ketone.
Ketones reacts with Grignard reagent to form tertiary alcohol
Butanal is an aldehyde.
Aldeyde reacts with grignard reagent to form Secondary alcohol
The answer is; Tertiary alcohol; Secondary alcohol