Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.
Answer: the line Spectra of hydrogen lies between the ultra-violet, visible light and infra-red of the electro magnetic spectrum
Explanation:
Electromagnetic radiation spans an wide range of wavelengths and frequencies. This range is called the electromagnetic spectrum. The electromagnetic spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The 7 regions includes; radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.
lower-energy radiation, such as radio waves, is expressed as frequency while microwaves, infrared, visible and UV light are usually expressed as wavelength and finally, higher-energy radiation such as X-rays and gamma rays, is expressed in terms of energy per photon.
Therefore, hydrogen lies between the ultra-violet, visible light and infra-red region of the electro magnetic spectrum.
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer:
<h3>Density of the Gas</h3>
Explanation:
More molecules mean more hits against the container walls. Increasing the number of particles means you have increased the density of the gas. This third factor is part of the ideal gas law, which explains how these three factors -- temperature, volume and density -- interact with each other.
Answer:
Ya'll should really practice more because things like these are easy
Explanation:
