On oxidation of aldehydes produces carboxylic acid functional group.
The product of oxidation of aldehydes depends upon whether the reaction occur in acidic medium or alkaline condition.
If oxidation of aldehydes occurs under acidic condition the product is carboxylic acid but if oxidation of aldehydes occurs under alkaline condition then reduce as well as oxidized product obtained which is known as disproportional product.
The oxidation of aldehydes occur through potassium dichromate, potassium permanganate or many more. The oxidation of aldehydes in the presence of base is known as cannizzaro's reaction.
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Hmm the answer is 2.454
No explanation needed it’s just 2.454
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
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Nonane (b) has the highest melting point.
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A caveat: I'm assuming that we're dealing with the straight-chain isomers of these alkanes (specifically pentane and nonane). The straight-chain isomer of pentane (<em>n</em>-pentane, CH3-[CH2]3-CH3) has a melting point of -129.8 °C; the straight-chain isomer of nonane (<em>n-</em>nonane, CH3-[CH2]7-CH3) has a melting point of -53.5 °C. The pattern holds as you go down (or up): The more carbon atoms, the higher the melting point. So, in decreasing order of melting points here, you'd have the following: nonane > pentane > butane > ethane.
However, one structural isomer of pentane, neopentane, has a melting point of -16.4 °C, which is <em>higher </em>that the melting point of <em>n</em>-nonane despite neopentane having the same molecular formula as its straight-chain isomer. Of course, you're not to blame for coming up with this question; this is just some extra info to keep in mind.
